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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 #311 : Calculus Ii

Evaluate the limit:

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

Possible Answers:

$\infty$

h^3-2h^2+3h-10

$-\infty$

DNE

Correct answer:

h^3-2h^2+3h-10

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:

$\lim_{x \to h}x^3-2x^2+3x-10=h^3-2h^2+3h-10$

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Example Question #312 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 4}x^2-8x+16$

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:

$\lim_{x \to 4}x^2-8x+16=4^2-8*4+16=16-32+16=32-32=0$

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Example Question #313 : Calculus Ii

Evaluate the limit:

$\lim_{x \to ab}x-ab+2a^2b^2$

Possible Answers:

$-\infty$

$\infty$

DNE

2a^2b^2

Correct answer:

2a^2b^2

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:

$\lim_{x \to ab}x-ab+2a^2b^2=ab-ab+2a^2b^2=2a^2b^2$

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Example Question #314 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 6}5x+6$

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:

$\lim_{x \to 6}5x+6=5*6+6=30+6=36$

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Example Question #315 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 10}x^2-2x+80$

Possible Answers:

160

$\infty$

$-\infty$

DNE

Correct answer:

160

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:

$\lim_{x \to 10}x^2-2x+80=10^2-2*10+80=100-20+80=80+80=160$

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Example Question #316 : Calculus Ii

Evaluate the limit:

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

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

$\lim_{x \to h}\frac{x^2-h^2}{x+h}=\frac{h^2-h^2}{h+h}=\frac{0}{2h}=0$

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Example Question #317 : Limits

Evaluate the limit:

$\lim_{x \to 5}\frac{x^2+2x+10}{x+5}$

Possible Answers:

DNE

4.5

$-\infty$

$\infty$

Correct answer:

4.5

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=5; so we proceed to insert the value of x into the entire equation.

$\lim_{x \to 5}\frac{x^2+2x+10}{x+5}=\frac{5^2+2*5+10}{5+5}=\frac{25+10+10}{10}=\frac{45}{10}=4.5$

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Example Question #317 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 5}\sqrt{x^2-2x+1}$

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:

$\lim_{x \to 5}\sqrt{x^2-2x+1}= \lim_{x \to 5}\sqrt{(x-1)^2}=\lim_{x \to 5}x-1=5-1=4$

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Example Question #318 : Calculus Ii

$\lim_{x \to 0}\sqrt{x^2+2x+2}$

Possible Answers:

DNE

$-\infty$

$\sqrt{2}$

$\infty$

Correct answer:

$\sqrt{2}$

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:

$\lim_{x \to 0}\sqrt{x^2+2x+2}=\sqrt{0^2+2*0+2}=\sqrt{2}$

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Example Question #319 : Calculus Ii

Evaluate the limit:

$\lim_{x \to 1}(x+2)^3$

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:

$\lim_{x \to 1}(x+2)^3=(1+2)^3=3^3=27$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #311 : Calculus Ii

Example Question #312 : Calculus Ii

Example Question #313 : Calculus Ii

Example Question #314 : Calculus Ii

Example Question #315 : Calculus Ii

Example Question #316 : Calculus Ii

Example Question #317 : Limits

Example Question #317 : Calculus Ii

Example Question #318 : Calculus Ii

Example Question #319 : Calculus Ii

All Calculus 2 Resources

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