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Calculus 2 : Limits

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Example Questions

Example Question #368 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{11x - 11}{-9sin(17\cdot \pi x)}\end{align*}$

Possible Answers:

$-\frac{11}{(153\cdot \pi )}$

$\infty$

$-\infty$

$\frac{11}{(153\cdot \pi )}$

Correct answer:

$\frac{11}{(153\cdot \pi )}$

Explanation:

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Example Question #411 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\end{align*}$

Possible Answers:

$\frac{(83521\cdot \pi ^{3})}{(\frac{(539\cdot \pi ^{2})}{2})^{(\frac{3}{2})}}$

$\infty$

$-\infty$

$-\frac{(83521\cdot \pi ^{3})}{(\frac{(539\cdot \pi ^{2})}{2})^{(\frac{3}{2})}}$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{-867\cdot \pi cos(17\cdot \pi x)sin(17\cdot \pi x)^{2}}{-154\cdot \pi cos(\frac{(7\cdot \pi x)}{2})sin(\frac{(7\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{14739\cdot \pi ^{2}sin(17\cdot \pi x)^{3} - 29478\cdot \pi ^{2}cos(17\cdot \pi x)^{2}sin(17\cdot \pi x)}{539\cdot \pi ^{2}sin(\frac{(7\cdot \pi x)}{2})^{2} - 539\cdot \pi ^{2}cos(\frac{(7\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{539\cdot \pi ^{2}}=0\\&\lim_{x\rightarrow1-}\frac{-17sin(17\cdot \pi x)^{3}}{22cos(\frac{(7\cdot \pi x)}{2})^{2}}=0\end{align*}$

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Example Question #412 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1+}\frac{49ln(x^{3})}{27x - 27}\end{align*}$

Possible Answers:

$\frac{49}{9}$

$-\frac{49}{9}$

$-\infty$

$\infty$

Correct answer:

$\frac{49}{9}$

Explanation:

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Example Question #371 : Finding Limits And One Sided Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&\lim_{x\rightarrow1-}\frac{5\cdot (x - 1)^{2}}{2cos(\frac{(19\cdot \pi x)}{2})^{2}}\end{align*}$

Possible Answers:

$-\frac{10}{(361\cdot \pi ^{2})}$

$-\infty$

$\frac{10}{(361\cdot \pi ^{2})}$

$\infty$

Correct answer:

$\frac{10}{(361\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{5\cdot (x - 1)^{2}}{2cos(\frac{(19\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{10x - 10}{-38\cdot \pi cos(\frac{(19\cdot \pi x)}{2})sin(\frac{(19\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{10}{361\cdot \pi ^{2}sin(\frac{(19\cdot \pi x)}{2})^{2} - 361\cdot \pi ^{2}cos(\frac{(19\cdot \pi x)}{2})^{2}}\rightarrow\frac{10}{361\cdot \pi ^{2}}\\&\lim_{x\rightarrow1-}\frac{5\cdot (x - 1)^{2}}{2cos(\frac{(19\cdot \pi x)}{2})^{2}}=\frac{10}{(361\cdot \pi ^{2})}\end{align*}$

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Example Question #371 : Finding Limits And One Sided Limits

Evaluate: $\lim_{x->3} \frac {x^4-81}{\sqrt {x^2-9}}$

Possible Answers:

Correct answer:

Explanation:

Screen shot 2016 01 01 at 9.27.10 pm

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Example Question #373 : Finding Limits And One Sided Limits

$\lim_{x->\infty}\frac{x^4+2x-3}{3x^4+x^2-2 }$

Possible Answers:

$\frac{1}{3}$

$\frac{3}{2}$

Does not exist

$\frac{\infty}{\infty}$

Correct answer:

$\frac{1}{3}$

Explanation:

If you chose $\frac{\infty}{\infty}$ this is incorrect. Although both the numerator and denominator become infinite when you apply the condition $x->\infty$ , the $\frac{\infty}{\infty}$ is one of several indeterminate forms, meaning we cannot conclude anything about the behavior of the function as $x->\infty$ with the expression in this form.

Solution

$\lim_{x->\infty}\frac{x^4+2x-3}{3x^4+x^2-2 }$

$=\lim_{x->\infty}\frac{\frac{1}{x^4}}{\frac{1}{x^4}}\frac{(x^4+2x-3)}{(3x^4+x^2-2)}$

$=\lim_{x->\infty}\frac{(1+\frac{2}{x^3}-\frac{3}{x^4})}{(3+\frac{1}{x^2}-\frac{2}{x^4})}=\frac{1+0-0}{3+0-0}$

The terms in the numerator and the denominator will all vanish in the limit except for in the numerator and in the denominator. This means the limit is $\frac{1}{3}$ .

Alternate Solution 1

We could also have used L'Hoptitals rule which allows us to take derivatives of the numerator and denominator,

$\lim_{x->\infty}\frac{f(x)}{g(x)}=\lim_{x->\infty}\frac{f'(x)}{g'(x)}$ .

This operation can be performed successively until the limit is in determinant form.

$L=\lim_{x->\infty}\frac{(x^4+2x-3)'}{(3x^4+x^2-2)' }$

$=\lim_{x->\infty}\frac{(4x^3+2)'}{(12x^3+2x)' }$

$=\lim_{x->\infty}\frac{(12x^2)'}{(36x^2+2)' }$

$=\lim_{x->\infty}\frac{(24x)'}{(72x)' }$

$=\lim_{x->\infty}\frac{24}{72}$

$=\frac{1}{3}$

Alternate Solution 2 (The Quickest!)

Notice that the coefficients in front of the highest order polynomial terms in the numerator and denominator can be read off immediately to write the limit. This approach works when the highest order terms are raised to the same power in the numerator and denominator.

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Example Question #374 : Finding Limits And One Sided Limits

Evaluate:

$\lim_{x\rightarrow 4+}\ln(x-4)-x^2$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

To evaluate the limit, we first must determine whether the limit is right or left sided; the positive sign exponent on 4 indicates that we are approaching 4 from the right side, using values slightly greater than 4. When we evaluate the limit, we see that the natural logarithm function goes to negative infinity as it approaches zero, and this dominates the behavior of the function overall (the polynomial is far smaller than infinity). So, our answer is $-\infty$ .

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Example Question #372 : Finding Limits And One Sided Limits

Evaluate:

$\lim_{x\rightarrow \infty}\frac{11x^2+10x}{x+4}$

Possible Answers:

$\frac{5}{2}$

$\infty$

$-\infty$

Correct answer:

$\infty$

Explanation:

To evaluate the limit, we must factor out a term consisting of the highest power term divided by itself (so we aren't changing the function itself):

$\lim_{x\rightarrow \infty} (\frac{x^2}{x^2})\frac{11+10x^{-1}}{x^{-1}+4x^{-2}}$

All of the terms with negative exponents equal zero as the x approaches infinity (they are all fractions with infinitely large denominators), so we are left with our function with zero in its denominator, which equals infinity.

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Example Question #373 : Finding Limits And One Sided Limits

Evaluate:

$\lim_{x\rightarrow \infty} \frac{x^3+x^2+x}{3^x+1}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

When evaluating the limit, we must look at the terms in the numerator and denominator. In the numerator, we have only polynomial terms. In the denominator, we have an exponential term, which grows faster than the terms in the numerator. Thus, the denominator grows faster than the numerator, so as x approaches infinity, the function goes to 0.

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Example Question #374 : Finding Limits And One Sided Limits

Evaluate:

$\lim_{x\rightarrow \infty} \frac{2x}{5x+1}$

Possible Answers:

$\frac{2}{5}$

$\infty$

$-\infty$

Correct answer:

$\frac{2}{5}$

Explanation:

To evaluate the limit, we must factor out a term consisting of the highest power term divided by itself (so we aren't changing the function itself):

$\lim_{x\rightarrow \infty} (\frac{x}{x})(\frac{2}{5+x^{-1}})$

The factor we created cancels to become one, and the negative exponent term goes to zero (the fraction has an infinitely large denominator). What we are left with is $\frac{2}{5}$ .

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #368 : Finding Limits And One Sided Limits

Example Question #411 : Limits

Example Question #412 : Limits

Example Question #371 : Finding Limits And One Sided Limits

Example Question #371 : Finding Limits And One Sided Limits

Example Question #373 : Finding Limits And One Sided Limits

Example Question #374 : Finding Limits And One Sided Limits

Example Question #372 : Finding Limits And One Sided Limits

Example Question #373 : Finding Limits And One Sided Limits

Example Question #374 : Finding Limits And One Sided Limits

All Calculus 2 Resources

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