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Calculus 2 : Calculus II

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

Example Questions

Example Question #41 : Limits

Find the limit.

$\lim_{t \to 0}\frac{\sqrt{t^2+9}-3}{t^2}$

Possible Answers:

$\frac{1}{6}$

$\infty$

Correct answer:

$\frac{1}{6}$

Explanation:

By plugging in 0, we get

$\frac{\sqrt{0^2+9}-3}{0^2}=\frac{0}{0}$ .

This solution is "indeterminate." Instead of directly plugging in 0, we simplify the function first, so we multiply by the conjugate of the numerator:

$\frac{\sqrt{t^2+9}-3}{t^2}*\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}$ .

Multiplying the numerators (via the FOIL method) and the denominators, we get

$\\\frac{(\sqrt{t^2+9})^2+3\sqrt{t^2+9}-3\sqrt{t^2+9}-9}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{t^2}{t^2(\sqrt{t^2+9}+3)}\\\\=\frac{1}{\sqrt{t^2+9}+3}$

Plugging in 0, we get

$\frac{1}{\sqrt{0^2+9}+3}=\frac{1}{6}$

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Example Question #42 : Limits

Find the value of the limit (if it exists):

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

Possible Answers:

Does not exist.

$\frac{2}{3}$

$\frac{-2}{3}$

Correct answer:

$\frac{2}{3}$

Explanation:

By "plugging in" 2, we get $\frac{0}{0}$ , which is indeterminate.

We need to manipulate the function to see if anything cancels out first.

By factoring, we have

$\lim_{x \to 2}\frac{x(x-2)}{(x+1)(x-2)}=\lim_{x \to 2}\frac{x}{(x+1)}=\frac{2}{3}$ .

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Example Question #43 : Limits

Evaluate

$\lim_{x \to 0}\frac{\sqrt{x+4}-2}{x}$

Possible Answers:

$\frac{-1}{4}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

Simplify the function first, then "plug in" 0 to evaluate the limit.

$\\\lim_{x \to 0}\frac{(\sqrt{x+4}-2)}{x}\frac{(\sqrt{x+4}+2)}{(\sqrt{x+4}+2)}\\\\=\lim_{x \to 0}\frac{(x+4)-4}{x(\sqrt{x+4}+2)}\\\\=\lim_{x \to 0}\frac{1}{\sqrt{x+4}+2}\\\\=\frac{1}{4}$

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Example Question #1 : Finding Limits And One Sided Limits

Which graph is a possible sketch of the function that possesses the following characteristics:

$f(-2)= undefined; \lim_{x \rightarrow -2^-} f(x)= \infty; \lim_{x \rightarrow -2^+} f(x)= -\infty;$

Possible Answers:

Graph2

Graph3

There does not exist such a graph.

Graph4

Graph1

Correct answer:

Graph1

Explanation:

Since $\tiny { f(-2) = undefined}$ there is a possible vertical asymptote at x=-2 .

As we approach x=-2 from the left, the graph should tend to $\tiny { +\infty}$ . Approaching x=-2 from the right, the graph tends to $\tiny {-\infty}$ .

The only graph that does so is Graph1 .

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Example Question #1 : Finding Limits And One Sided Limits

Rational_graph

The graph above is a sketch of the function . Find $\lim_{x \rightarrow 2^-}f(x)$ .

Possible Answers:

Does not exist

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

We need to look at the behavior of the function as it tends to from the left. Therefore the answer is $\tiny { -\infty}$ .

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Example Question #1 : Finding Limits And One Sided Limits

Rational_graph

The graph above is a sketch of . Find $\lim_{x \rightarrow -2} f(x)$ .

Possible Answers:

$-\infty$

Does not exist

$\infty$

Correct answer:

Does not exist

Explanation:

The limit does not exist because the one-sided limits are not equal; $\lim_{x \rightarrow -2^-}f(x) = -\infty$ , whereas $\lim_{x \rightarrow -2^+}f(x) = +\infty$ .

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Example Question #3 : Finding Limits And One Sided Limits

Determine the value of $\lim_{x \to 5^-} \left |-5 \right |$ .

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

The expression ${}\lim_{x \to 5^-} \left |-5 \right |$ indicates that all points on the domain are equal to 5 since the absolute value negates negative values.

The $\lim_{x \to 5^-}$ is to search for the limit as the graph approaches x=5 from the left side of the graph. Since the absolute value of negative five is five, the graph approaches five from the left.

The correct answer is .

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Example Question #2 : Finding Limits And One Sided Limits

Evaluate the following limit.

$\small \lim_{x\to 9} \frac{x-9}{\sqrt{x}-3}$

Possible Answers:

Limit does not exist

$\small 6$

$\small 3+\sqrt{3}$

Correct answer:

$\small 6$

Explanation:

This limit can be solved using simple manipulation of the expression inside the limit:

$\\ \small \small \lim_{x\to 9} \frac{x-9}{\sqrt{x}-3}\\ \\=\lim_{x\to 9} \frac{x-9}{\sqrt{x}-3}\cdot \frac{\sqrt{x}+3}{\sqrt{x}+3}\\ \\=\lim_{x\to 9} \frac{x-9}{x-9}\cdot (\sqrt{x}+3)\\ \\=\lim_{x\to 9} \sqrt{x}+3\\ \\=\sqrt{9}+3=6$

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Example Question #3 : Finding Limits And One Sided Limits

Evaluatle the limit:

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

Possible Answers:

DNE

Correct answer:

Explanation:

Consider the domain of the function. Because this equation is a polynomial, x is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that x is approaching into the limit equation.

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

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Example Question #4 : Finding Limits And One Sided Limits

Evaluate the limit:

$\lim_{x \to -2}\sqrt{x^4+3x+6}$

Possible Answers:

DNE

Correct answer:

Explanation:

$\lim_{x \to -2}\sqrt{x^4+3x+6}=\sqrt{(-2)^4+3(-2)+6}=\sqrt{16-6+6}=\sqrt{16}=4$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Limits

Example Question #42 : Limits

Example Question #43 : Limits

Example Question #1 : Finding Limits And One Sided Limits

Example Question #1 : Finding Limits And One Sided Limits

Example Question #1 : Finding Limits And One Sided Limits

Example Question #3 : Finding Limits And One Sided Limits

Example Question #2 : Finding Limits And One Sided Limits

Example Question #3 : Finding Limits And One Sided Limits

Example Question #4 : Finding Limits And One Sided Limits

All Calculus 2 Resources

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