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Calculus 2 : Limits

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Example Questions

Example Question #431 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\end{align*}$

Possible Answers:

$\infty$

$-\frac{(35\cdot \pi )}{2}$

$-\infty$

$\frac{(35\cdot \pi )}{2}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-140\cdot \pi cos(7\cdot \pi x)}{\frac{(24ln(x^{8})^{2})}{x}}\rightarrow\frac{140\cdot \pi}{0}=\infty\\&lim_{x\rightarrow1+}\frac{-20sin(7\cdot \pi x)}{ln(x^{8})^{3}}=\infty\end{align*}$

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Example Question #432 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{10x}{19ln((x + 1)^{8})^{2}}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$-\frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$

$\frac{(5\cdot 1216^{(\frac{1}{2})})}{608}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{10x}{19ln((x + 1)^{8})^{2}}\rightarrow\frac{0}{0}\\&\frac{10}{\frac{(304ln((x + 1)^{8}))}{(x + 1)}}\rightarrow\frac{10}{0}=\infty\\&lim_{x\rightarrow0+}\frac{10x}{19ln((x + 1)^{8})^{2}}=\infty\end{align*}$

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Example Question #433 : Limits

$\lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}$

Possible Answers:

2xh

Correct answer:

Explanation:

This question is asking you to recall the limit definition of a derivative, which states:

$\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}=f'(x).$

Therefore, if we can define our function for our problem, we can simply evaluate this limit by taking its derivative.

$\lim_{h\rightarrow 0} \frac{(x+h)^2 - x^2}{h}.$ In these types of questions, you look at the second term in the numerator. That is the negative of the function.

Since f(x)=x^2 , f'(x)=2x.

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Example Question #434 : Limits

Evaluate the following limit:

$\lim_{x \to 0} \frac{2x^2}{5}$

Possible Answers:

$\frac{2}{5}$

$\infty$

$-\infty$

Correct answer:

Explanation:

The first step is to always plug in the value of the limit. Doing so we get

$\frac{2(0)^2)}{5}=0$

Since this is a valid answer, no further work is required.

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Example Question #435 : Limits

Evaluate the following limit:

$\lim_{x \to 0} \frac{5x+10}{x+1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

The first step is to always plug in the value of the limit. Doing so we get

$\frac{5(0)+10}{0+1}=\frac{10}{1}=10$

Since this is a valid answer, no further work is required.

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Example Question #436 : Limits

Evaluate the following limit:

$\lim_{x \to 2} \frac{3x^2 + 2x}{x-1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

The first step is to always plug in the value of the limit. Doing so we get

$\frac{3(2)^2+2(2)}{2-1}=\frac{(3)(4)+(2)(2)}{1}=16$

Since this is a valid answer, no further work is required.

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Example Question #434 : Limits

Evaluate the following limit:

$\lim_{x \to 3} \frac{x^2-x-6}{x^2-9}$

Possible Answers:

$-\infty$

$\textup{Indeterminate}$

$\frac{5}{6}$

Correct answer:

$\frac{5}{6}$

Explanation:

The first step is to always plug in the value of the limit. Doing so we get

$\frac{(3^2-3-6)}{3^2-9}=\frac{0}{0}$

This is indeterminate. We need to verify it is indeterminate and therefore need take another approach. Let's take a look at the original equation again.

$\frac{x^2-x-6}{x^2-9}$

Taking what we learned in algebra, we know that both the numerator and denominator of the equation are factorable. The equation becomes

$\frac{(x-3)(x+2)}{(x-3)(x+3)}$

The (x-3) from the top and bottom cancels, which then simplifies to

$\frac{(x+2)}{(x+3)}$

Now we evaluate the limit once again using the simplified equation and we get

$\frac{(3+2)}{(3+3)}=\frac{5}{6}$

This is a valid answer.

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Example Question #435 : Limits

$\lim_{x \to 0} \frac{(4x^2-2x)}{(x)}$

Possible Answers:

$\textup{Indeterminate}$

$\infty$

Correct answer:

Explanation:

The first step is to always plug in the value of the limit. Doing so we get

$\frac{(4(0)^2-2(0))}{0}=\frac{0}{0}$

This is indeterminate. We need to verify it is indeterminate and therefore need take another approach. Let's take a look at the original equation again.

$\frac{4x^2-2x}{x}$

Taking what we learned in algebra, we know that the numerator is factorable. Taking out an yields

$\frac{(x)(4x-2)}{(x)}$

The (x) from the top and bottom cancels, which then simplifies to

$\frac{(4x-2)}{(1)}$

Now we evaluate the limit once again using the simplified equation and we get

4(0)-2=-2

This is a valid answer

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Example Question #394 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\end{align*}$

Possible Answers:

$-\frac{41}{135424}$

$\infty$

$-\infty$

$\frac{41}{135424}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}\rightarrow\frac{0}{0}\\&\frac{82x - 82}{\frac{368}{x}}\rightarrow\frac{0}{368}=0\\&lim_{x\rightarrow1-}\frac{41\cdot (x - 1)^{2}}{46ln(x^{8})}=0\end{align*}$

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Example Question #437 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}\frac{49ln(x^{8})}{48\cdot (x - 1)^{3}}\end{align*}$

Possible Answers:

$\frac{(49\cdot (-48)^{(\frac{2}{3})})}{6}$

$\infty$

$-\infty$

$-\frac{(49\cdot (-48)^{(\frac{2}{3})})}{6}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{49ln(x^{8})}{48\cdot (x - 1)^{3}}\rightarrow\frac{0}{0}\\&\frac{\frac{392}{x}}{144\cdot (x - 1)^{2}}\rightarrow\frac{392}{0}=\infty\\&lim_{x\rightarrow1-}\frac{49ln(x^{8})}{48\cdot (x - 1)^{3}}=\infty\end{align*}$

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #431 : Limits

Example Question #432 : Limits

Example Question #433 : Limits

Example Question #434 : Limits

Example Question #435 : Limits

Example Question #436 : Limits

Example Question #434 : Limits

Example Question #435 : Limits

Example Question #394 : Finding Limits And One Sided Limits

Example Question #437 : Limits

All Calculus 2 Resources

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