We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$\frac{12}{7}$ $\displaystyle \frac{12}{7}$

$\infty$ $\displaystyle \infty$

$-\frac{12}{7}$ $\displaystyle -\frac{12}{7}$

Correct answer:

$\frac{12}{7}$ $\displaystyle \frac{12}{7}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(2\cdot (x - 8)\cdot (x - 2)\cdot (x - 7))}{(7\cdot (x - 8))}\\&\frac{(2\cdot (x - 2)\cdot (x - 7))}{7}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}=\frac{12}{7}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(6x^{2} - 68x + 172)}{(7)}=\frac{12}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(2\cdot (x - 8)\cdot (x - 2)\cdot (x - 7))}{(7\cdot (x - 8))}\\&\frac{(2\cdot (x - 2)\cdot (x - 7))}{7}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}=\frac{12}{7}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(6x^{2} - 68x + 172)}{(7)}=\frac{12}{7}\end{align*}$

Report an Error

Example Question #1031 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$\frac{8}{49}$ $\displaystyle \frac{8}{49}$

$\infty$ $\displaystyle \infty$

$-\frac{8}{49}$ $\displaystyle -\frac{8}{49}$

Correct answer:

$-\frac{8}{49}$ $\displaystyle -\frac{8}{49}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(16\cdot (x + 3)\cdot (x - 2)\cdot (x + 5))}{((x - 2)\cdot (x + 3)\cdot (x - 4)\cdot (x + 7)\cdot (x + 10))}\\&\frac{(16\cdot (x + 5))}{((x - 4)\cdot (x + 7)\cdot (x + 10))}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}=-\frac{8}{49}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-3-}\frac{(192x + 48x^{2} - 16)}{(27x^{2} - 712x + 56x^{3} + 5x^{4} - 292)}=-\frac{8}{49}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(16\cdot (x + 3)\cdot (x - 2)\cdot (x + 5))}{((x - 2)\cdot (x + 3)\cdot (x - 4)\cdot (x + 7)\cdot (x + 10))}\\&\frac{(16\cdot (x + 5))}{((x - 4)\cdot (x + 7)\cdot (x + 10))}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}=-\frac{8}{49}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-3-}\frac{(192x + 48x^{2} - 16)}{(27x^{2} - 712x + 56x^{3} + 5x^{4} - 292)}=-\frac{8}{49}\end{align*}$

Report an Error

Example Question #71 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\frac{63}{4}$ $\displaystyle \frac{63}{4}$

$-\frac{63}{4}$ $\displaystyle -\frac{63}{4}$

Correct answer:

$-\frac{63}{4}$ $\displaystyle -\frac{63}{4}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x + 8)\cdot (x + 1)\cdot (x - 5)\cdot (x - 10))}{(6\cdot (x + 8)\cdot (x - 4)\cdot (x - 5))}\\&\frac{(3\cdot (x + 1)\cdot (x - 10))}{(2\cdot (x - 4))}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}=-\frac{63}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8-}\frac{(36x^{3} - 162x^{2} - 1386x + 2970)}{(18x^{2} - 12x - 312)}=-\frac{63}{4}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x + 8)\cdot (x + 1)\cdot (x - 5)\cdot (x - 10))}{(6\cdot (x + 8)\cdot (x - 4)\cdot (x - 5))}\\&\frac{(3\cdot (x + 1)\cdot (x - 10))}{(2\cdot (x - 4))}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}=-\frac{63}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8-}\frac{(36x^{3} - 162x^{2} - 1386x + 2970)}{(18x^{2} - 12x - 312)}=-\frac{63}{4}\end{align*}$

Report an Error

Example Question #74 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}\end{align*}$

Possible Answers:

$-\frac{8}{9}$ $\displaystyle -\frac{8}{9}$

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

Correct answer:

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(8x - 64)}{(7x - x^{2} + 8)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(8\cdot (x - 8))}{((x + 1)\cdot (x - 8))}\\&\frac{8}{(x + 1)}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}=\frac{8}{9}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(8)}{(2x - 7)}=\frac{8}{9}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(8x - 64)}{(7x - x^{2} + 8)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(8\cdot (x - 8))}{((x + 1)\cdot (x - 8))}\\&\frac{8}{(x + 1)}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}=\frac{8}{9}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(8)}{(2x - 7)}=\frac{8}{9}\end{align*}$

Report an Error

Example Question #72 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\frac{ 73}{82}$ $\displaystyle \frac{ 73}{82}$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$\displaystyle 0$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x - 9)^{2})}{(17\cdot (x - 9)\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&\frac{(4\cdot (x - 9))}{(17\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow9+}\frac{(8x - 72)}{(9928x - 5559x^{2} - 272x^{3} + 85x^{4} + 38556)}=0\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x - 9)^{2})}{(17\cdot (x - 9)\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&\frac{(4\cdot (x - 9))}{(17\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow9+}\frac{(8x - 72)}{(9928x - 5559x^{2} - 272x^{3} + 85x^{4} + 38556)}=0\end{align*}$

Report an Error

Example Question #73 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\end{align*}$

Possible Answers:

$-\frac{86}{69}$ $\displaystyle -\frac{86}{69}$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-4\text{, as that'd}\\&\text{create a zero value in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\\&\text{can be factored. This'll allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(17\cdot (x + 4)\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x + 4)^{2}\cdot (x - 3)\cdot (x + 5))}\\&\frac{(17\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x - 3)\cdot (x + 4)\cdot (x + 5))}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}=\infty\\&\text{Another method, since numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4-}\frac{(102x + 306x^{2} + 68x^{3} - 442)}{(714x^{2} - 2464x + 560x^{3} + 70x^{4} - 3360)}=\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-4\text{, as that'd}\\&\text{create a zero value in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\\&\text{can be factored. This'll allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(17\cdot (x + 4)\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x + 4)^{2}\cdot (x - 3)\cdot (x + 5))}\\&\frac{(17\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x - 3)\cdot (x + 4)\cdot (x + 5))}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}=\infty\\&\text{Another method, since numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4-}\frac{(102x + 306x^{2} + 68x^{3} - 442)}{(714x^{2} - 2464x + 560x^{3} + 70x^{4} - 3360)}=\infty\end{align*}$

Report an Error

Example Question #71 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\end{align*}$

Possible Answers:

$-\frac{304}{117}$ $\displaystyle -\frac{304}{117}$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

Correct answer:

$-\frac{304}{117}$ $\displaystyle -\frac{304}{117}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(19\cdot (x - 2)\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x - 2)\cdot (x + 2)\cdot (x + 7))}\\&\frac{(19\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x + 2)\cdot (x + 7))}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}=-\frac{304}{117}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow2+}\frac{(76x^{3} - 399x^{2} - 1748x + 3268)}{(182x + 39x^{2} - 52)}=-\frac{304}{117}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(19\cdot (x - 2)\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x - 2)\cdot (x + 2)\cdot (x + 7))}\\&\frac{(19\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x + 2)\cdot (x + 7))}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}=-\frac{304}{117}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow2+}\frac{(76x^{3} - 399x^{2} - 1748x + 3268)}{(182x + 39x^{2} - 52)}=-\frac{304}{117}\end{align*}$

Report an Error

Example Question #71 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\frac{40}{3}$ $\displaystyle \frac{40}{3}$

$-\frac{40}{3}$ $\displaystyle -\frac{40}{3}$

Correct answer:

$-\frac{40}{3}$ $\displaystyle -\frac{40}{3}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(10x^{2}\cdot (x + 2)\cdot (x + 1)\cdot (x + 8))}{(18\cdot (x + 2))}\\&\frac{(5x^{2}\cdot (x + 1)\cdot (x + 8))}{9}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}=-\frac{40}{3}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-2+}\frac{(320x + 780x^{2} + 440x^{3} + 50x^{4})}{(18)}=-\frac{40}{3}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(10x^{2}\cdot (x + 2)\cdot (x + 1)\cdot (x + 8))}{(18\cdot (x + 2))}\\&\frac{(5x^{2}\cdot (x + 1)\cdot (x + 8))}{9}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}=-\frac{40}{3}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-2+}\frac{(320x + 780x^{2} + 440x^{3} + 50x^{4})}{(18)}=-\frac{40}{3}\end{align*}$

Report an Error

Example Question #72 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$-\frac{1}{105}$ $\displaystyle -\frac{1}{105}$

$\frac{1}{105}$ $\displaystyle \frac{1}{105}$

Correct answer:

$-\frac{1}{105}$ $\displaystyle -\frac{1}{105}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(2x\cdot (x - 5))}{(7x\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&\frac{(2\cdot (x - 5))}{(7\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}=-\frac{1}{105}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(4x - 10)}{(1330x + 378x^{2} + 28x^{3} + 1050)}=-\frac{1}{105}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(2x\cdot (x - 5))}{(7x\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&\frac{(2\cdot (x - 5))}{(7\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}=-\frac{1}{105}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(4x - 10)}{(1330x + 378x^{2} + 28x^{3} + 1050)}=-\frac{1}{105}\end{align*}$

Report an Error

Example Question #81 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$-\frac{3}{4}$ $\displaystyle -\frac{3}{4}$

$\infty$ $\displaystyle \infty$

Correct answer:

$-\frac{3}{4}$ $\displaystyle -\frac{3}{4}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x + 7)\cdot (x - 8))}{(2\cdot (x + 7)\cdot (x - 3)\cdot (x - 5))}\\&\frac{(6\cdot (x - 8))}{((x - 3)\cdot (x - 5))}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}=-\frac{3}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7+}\frac{(24x - 12)}{(6x^{2} - 4x - 82)}=-\frac{3}{4}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x + 7)\cdot (x - 8))}{(2\cdot (x + 7)\cdot (x - 3)\cdot (x - 5))}\\&\frac{(6\cdot (x - 8))}{((x - 3)\cdot (x - 5))}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}=-\frac{3}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7+}\frac{(24x - 12)}{(6x^{2} - 4x - 82)}=-\frac{3}{4}\end{align*}$

Report an Error

View Calculus 3 Tutors

Peter
Certified Tutor

Kansas State University, Bachelors, Mathematics. Bowling Green State University-Main Campus, Master of Science, Management In...

View Calculus 3 Tutors

Nicholas
Certified Tutor

Bryant University, Bachelor of Science, Actuarial Science.

View Calculus 3 Tutors

Philip
Certified Tutor

Texas Tech University, Doctor of Philosophy, Higher Education Administration.

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Houston, French Tutors in San Francisco-Bay Area, Physics Tutors in San Diego, Spanish Tutors in Seattle, ACT Tutors in Los Angeles, ISEE Tutors in New York City, GMAT Tutors in New York City, Reading Tutors in New York City, Calculus Tutors in Denver, MCAT Tutors in Seattle

Popular Courses & Classes

GRE Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Houston, ACT Courses & Classes in New York City, GRE Courses & Classes in San Diego, Spanish Courses & Classes in Miami, ISEE Courses & Classes in Houston, ACT Courses & Classes in Seattle, Spanish Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in New York City, LSAT Courses & Classes in Atlanta

Popular Test Prep

GMAT Test Prep in Dallas Fort Worth, ACT Test Prep in Seattle, GRE Test Prep in Seattle, SAT Test Prep in New York City, MCAT Test Prep in Chicago, LSAT Test Prep in San Diego, SAT Test Prep in Atlanta, LSAT Test Prep in Seattle, GMAT Test Prep in Chicago, GRE Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Limits

Example Question #1031 : Calculus 3

Example Question #71 : Partial Derivatives

Example Question #74 : Partial Derivatives

Example Question #72 : Partial Derivatives

Example Question #73 : Partial Derivatives

Example Question #71 : Limits

Example Question #71 : Partial Derivatives

Example Question #72 : Partial Derivatives

Example Question #81 : Limits

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: