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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 #331 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 8}\sqrt{x^2+2x+1}$

Possible Answers:

$-\infty$

$\infty$

DNE

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:

$\lim_{x \to 8}\sqrt{x^2+2x+1}=\lim_{x \to 8}\sqrt{(x+1)^2}=\lim_{x \to 8}x+1=8+1=9$

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Example Question #332 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 9}5x-10+3x^2$

Possible Answers:

278

$-\infty$

DNE

$\infty$

Correct answer:

278

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:

$\lim_{x \to 9}5x-10+3x^2=5*9-10+3*9^2=45-10+243=35+243=278$

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Example Question #333 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 10}\frac{x^2-100}{x+10}$

Possible Answers:

$-\infty$

DNE

$\infty$

Correct answer:

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=10; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 10}\frac{x^2-100}{x+10}=\frac{10^2-100}{10+10}=\frac{100-100}{20}=\frac{0}{20}=0$

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Example Question #334 : Calculus Ii

Evaluate the limit:

$\lim_{x \to h}(x+2)^2$

Possible Answers:

$\infty$

DNE

$-\infty$

h^2+4h+4

Correct answer:

h^2+4h+4

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:

$\lim_{x \to h}(x+2)^2=(h+2)^2=h^2+4h+4$

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Example Question #335 : Calculus Ii

Evaluate the limit:

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

Possible Answers:

DNE

$\infty$

$-\infty$

Correct answer:

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=3; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 3}\frac{x^2-2x+1}{x-2}=\frac{3^2-2*3+1}{3-2}=\frac{9-6+1}{1}=\frac{3+1}{1}=4$

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Example Question #336 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 9}x^2-2x-1$

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:

$\lim_{x \to 9}x^2-2x-1=9^2-2*9-1=81-1-18=62$

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Example Question #337 : Calculus Ii

Evaluate the limit:

$\lim_{x \to \sqrt{3}}x^2$

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:

$\lim_{x \to \sqrt{3}}x^2=(\sqrt{3})^2=3$

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Example Question #338 : Calculus Ii

Evaluate the limit:

$\lim_{x \to t^2}x-20+t$

Possible Answers:

$\infty$

t^2-20+t

$-\infty$

DNE

Correct answer:

t^2-20+t

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:

$\lim_{x \to t^2}x-20+t=t^2-20+t$

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Example Question #339 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 0}5x-10$

Possible Answers:

$-\infty$

$\infty$

DNE

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:

$\lim_{x \to 0}5x-10=5*0-10=-10$

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Example Question #340 : 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:

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

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #331 : Calculus Ii

Example Question #332 : Calculus Ii

Example Question #333 : Calculus Ii

Example Question #334 : Calculus Ii

Example Question #335 : Calculus Ii

Example Question #336 : Calculus Ii

Example Question #337 : Calculus Ii

Example Question #338 : Calculus Ii

Example Question #339 : Calculus Ii

Example Question #340 : Calculus Ii

All Calculus 2 Resources

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