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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #341 : Calculus Ii

Evluate the limit:

$\lim_{x \to \pi}\sin(x)+\cos (x)$

Possible Answers:

$-\infty$

DNE

$\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to \pi}\sin(x)+\cos (x)=\sin (\pi)+\cos (\pi)=0+1=1$

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Example Question #342 : Calculus Ii

Evluate the limit:

$\lim_{x \to 1}\ln (x)-2$

Possible Answers:

DNE

$\infty$

$-\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to 1}\ln (x)-2=\ln(1)-2=0-2=-2$

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Example Question #343 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 100}\log_{10}(x)$

Possible Answers:

DNE

$\infty$

$-\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to 100}\log_{10}(x)=\log_{10}(100)=2$

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Example Question #344 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 3}\log_{x}(27)$

Possible Answers:

$-\infty$

DNE

$\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to 3}\log_{x}(27)=\log_{3}(27)=3$

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Example Question #345 : Calculus Ii

Evaluate the limit:

$\lim_{x \to \pi}\csc(x)\sec(x)$

Possible Answers:

DNE

$\infty$

$-\infty$

Correct answer:

DNE

Explanation:

The cosecant is the reciprical of the sine of an angle. The sine of pi is 0; so the reciprical of 0 is DNE. So when we insert x=pi into the equation, we get:

$\lim_{x \to \pi}\csc(x)\sec(x)=\csc(\pi)\sec(\pi)=\frac{1}{0}*1\Rightarrow DNE$

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Example Question #346 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 0}\frac{10x^2-20}{5x}$

Possible Answers:

DNE

$-\infty$

$\infty$

Correct answer:

DNE

Explanation:

We see that we cannot factor this to make the denominator not equal 0; hence this limit DNE because the denominator is zero.

$\lim_{x \to 0}\frac{10x^2-20}{5x}=\frac{10*0^2-20}{5*0}=\frac{-20}{0}\Rightarrow DNE$

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Example Question #347 : Calculus Ii

Evaluate the limit:

$\lim_{x \to \frac{\pi }{2}}\sin(x)\cos (x)$

Possible Answers:

DNE

$\infty$

$-\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{x \to \frac{\pi }{2}}\sin(x)\cos (x)=\sin (\frac{\pi }{2})\cos (\frac{\pi }{2})=1*0=0$

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Example Question #348 : Limits

Find the limit, approaching from the right:

$\lim_{x \to 0^+} \frac{\cos (x)}{x}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

To find the limit of the whole thing we have to find the individual limits of the top and bottom of the fraction, coming from the right.

For the top:

$\cos(x) \rightarrow 1$ as $x \rightarrow 0$ , because $\cos(0)=1$ and cos(x) is a continuous function.

$x \rightarrow 0$ as you approach 0.

So, near 0 we're looking at what's basically $\frac{1}{x}$ . And we know that $\cos(x)$ is positive for positive values of x near 0, so with these two facts combined we have enough information to see that

$\lim_{x \to 0^+} \frac{\cos x}{x} = \infty$ .

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Example Question #348 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}$

Possible Answers:

$-\infty$

$\frac{1}{6}$

Does not exist

$\infty$

Correct answer:

$\frac{1}{6}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 0}\frac{x^2+2x+1}{x+6}=\frac{0^2+2*0+1}{0+6}=\frac{1}{6}$

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Example Question #349 : Calculus Ii

Evaluate the following limit:

$\lim_{x\rightarrow \infty}\frac{x^4}{x^3+5x^2+x}$

Possible Answers:

$-\infty$

$\infty$

$\frac{1}{5}$

Correct answer:

$\infty$

Explanation:

To evaluate the limit, we must first pull out a factor consisting of the highest power term divided by itself (one, essentially):

$\lim_{x\rightarrow \infty} \frac{x^4}{x^4}(\frac{1}{x^{-1}+5x^{-2}+x^{-3}})$

After the factor we created becomes 1, the negative exponent terms go to zero as x approaches infinity, therefore we are left with $\frac{1}{0}=\infty$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #341 : Calculus Ii

Example Question #342 : Calculus Ii

Example Question #343 : Calculus Ii

Example Question #344 : Calculus Ii

Example Question #345 : Calculus Ii

Example Question #346 : Calculus Ii

Example Question #347 : Calculus Ii

Example Question #348 : Limits

Example Question #348 : Calculus Ii

Example Question #349 : Calculus Ii

All Calculus 2 Resources

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