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Calculus 2 : Limits

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Example Questions

Example Question #301 : Limits

What is the $\lim_{x \rightarrow \sqrt3^+} y$ , looking at the given graph of , $y=\frac{x+4}{x^2-3}$

Possible Answers:

$-\infty$

Does not exist

$\infty$

Correct answer:

$\infty$

Explanation:

As shown by the graph, the limit as goes to $\sqrt3$ from the right is infinity.

To make sure that the right limit is $\sqrt3$ , we can plug the number into the denominator of the function and see if the denominator equals .

$x^2-3 = (\sqrt3)^2-3 = 3-3 = 0$

The denominator is equal to , so we can say that limit is infinity

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Example Question #302 : Limits

What is the $\lim_{x \rightarrow 2^-} y$ , for $y=\frac{2x^3+4}{2x-4}$

Possible Answers:

$\infty$

$-\infty$

Does not exist

Correct answer:

$-\infty$

Explanation:

As you go to x=2 , the function will get closer and closer to having for the denominator, sending the total value of the function into infinity.

Just to make sure, plug in into the denominator so that it is equal to .

$2x-4 = 2\cdot2-4 = 4-4 = 0$

As we go to from the left, the function is tending towards negative infinity. It is a negative because when going to from the left, the function will always be less than , hence negative.

So the limit is $-\infty$

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Example Question #303 : Limits

What is the $\lim_{x \rightarrow -\frac{1}{5}^+} y$ , given the following graph of ,

$y=\frac{3x^2+x}{5x+1}$

Possible Answers:

$-\infty$

$\infty$

Does not exist

Correct answer:

$-\infty$

Explanation:

As shown by the graph, the limit as goes to $-\frac{1}{5}$ from the right is negative infinity.

To make sure that the right limit is $-\frac{1}{5}$ , we can plug the number into the denominator of the function and see if the denominator equals .

$5x+1 = -5\cdot\frac{1}{5}+1 = -1 + 1 = 0$

The denominator is equal to , so we can say that limit is negative infinity

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Example Question #304 : Limits

What is the $\lim_{x \rightarrow 6^+} y$ , given the following graph of ,

$y=\frac{x^2+4}{x-6}$

Possible Answers:

$-\infty$

$\infty$

Does not exist

Correct answer:

$\infty$

Explanation:

As shown by the graph, the limit as goes to from the right is infinity.

To make sure that the right limit is at , we can plug the number into the denominator of the function and see if the denominator equals .

x-6 = 6-6 = 0

The denominator is equal to , so we can say that limit is infinity.

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Example Question #305 : Limits

What is the $\lim_{x \rightarrow 3^+} y$ , given the following graph of ,

$y=\frac{x+4}{x^3-27}$

Possible Answers:

Does not exist

$\infty$

$-\infty$

Correct answer:

$\infty$

Explanation:

As shown by the graph, the limit as goes to from the right is infinity.

To make sure that the right limit is , we can plug the number into the denominator of the function and see if the denominator equals .

x^3-27 = 3^3 - 27 = 27-27 = 0

The denominator is equal to , so this limit will go to infinity as goes to from the right.

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Example Question #306 : Limits

Let f(x) be the piecewise function denoted below

$f(x)=\left\{\begin{matrix} 3& x<2\\ x^2& x\geq 2 \end{matrix}\right.$

Evaluate the limit

$\lim_{x\rightarrow 2^+}f(x)$

Possible Answers:

DNE

Correct answer:

Explanation:

The limit

$\lim_{x\rightarrow a^+}f(x)$

denotes the limit of the function f(x) as approaches from the right.

For $\lim_{x\rightarrow 2^+}f(x)$

We can evaluate $f(2.1), \,f(2.01)$ , and f(2.001) to find the limit.

Because $2.1,\, 2.01,\, 2.001 \geq 2$

f(2.1)=2.1^2=4.41

f(2.01)=2.01^2=4.0401

f(2.001)=2.001^2=4.004001

From this pattern we find the limit to be

$\lim_{x\rightarrow 2^+}f(x) =4$

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Example Question #307 : Limits

Let f(x) be the piecewise function denoted below

$f(x)=\left\{\begin{matrix} 4& x<0\\ e^x& x\geq 0 \end{matrix}\right.$

Evaluate the limit

$\lim_{x\rightarrow 0^-}f(x)$

Possible Answers:

DNE

Correct answer:

Explanation:

The limit

$\lim_{x\rightarrow a^-}f(x)$

denotes the limit of the function f(x) as approaches from the left.

For $\lim_{x\rightarrow 0^-}f(x)$

We can evaluate $f(-0.1), \,f(-0.01)$ , and f(-0.001) to find the limit.

Because $-0.1,\, -0.01,\, -0.001 < 0$

f(-0.1)=4

f(-0.01)=4

f(-0.001)=4

From this pattern we find the limit to be

$\lim_{x\rightarrow 0^-}f(x) =4$

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Example Question #308 : Limits

Let f(x) be the piecewise function denoted below

$f(x)=\left\{\begin{matrix} 1& x<0\\ e^x& x\geq 0 \end{matrix}\right.$

Evaluate the limit if it exists

$\lim_{x\rightarrow 0}f(x)$

Possible Answers:

DNE

$\infty$

Correct answer:

Explanation:

The limit

$\lim_{x\rightarrow 0}f(x)$

exists if

$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)$

Because

$\lim_{x\rightarrow 0^-}f(x)=1$

and

$\lim_{x\rightarrow 0^+}f(x)=e^0=1$

We find that $\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)$

and the limit exists.

As such

$\lim_{x\rightarrow 0}f(x)=1$

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Example Question #309 : Limits

Evaluate the following limit:

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

Possible Answers:

$-\infty$

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

While the numerator of the function is a larger degree polynomial than the denominator, the exponential function in the denominator grows faster than the polynomial. Therefore, as the limit approaches infinity, the denominator becomes far larger than the numerator, making the fraction go to .

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Example Question #310 : Limits

Evaluate the limit:

$\lim_{x \to 0}x^2+2x+4$

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:

$\lim_{x \to 0}x^2+2x+4=0^2+2*0+4=4$

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #301 : Limits

Example Question #302 : Limits

Example Question #303 : Limits

Example Question #304 : Limits

Example Question #305 : Limits

Example Question #306 : Limits

Example Question #307 : Limits

Example Question #308 : Limits

Example Question #309 : Limits

Example Question #310 : Limits

All Calculus 2 Resources

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