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Calculus 2 : Limits

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Example Questions

Example Question #351 : Limits

Evaluate the following limit:

$\lim_{x\rightarrow 4^+}f(x), f(x)=\left\{\begin{matrix} x^2+1, x\leq 4\\ \ln(x-4), x> 4 \end{matrix}\right.$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

To evaluate the limit, we must determine which side of the limit we are approaching. The plus sign "exponent" means that we are approaching 4 from the right, using numbers larger than 4. Therefore, we must evaluate the limit using the second part of the piecewise function; when we evaluate the limit, we get the limit as the natural logarithm approaches zero, which gives us $-\infty$ .

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Example Question #352 : Limits

Evaluate the following limit:

$\lim_{x\rightarrow 3} \frac{\ln(3-x)}{2}$

Possible Answers:

$-\frac{1}{3}$

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

To evaluate the limit, we simply substitute the x value we are approaching - 3 - into the limit. As the natural logarithm gets closer to zero, we reach $-\infty$ .

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Example Question #353 : Limits

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

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

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

By L'Hopital's rule:

$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}$

$\lim_{x\rightarrow \infty} \frac{4x}{x^2+5x}=\lim_{x\rightarrow \infty} \frac{4}{2x+5}=0$

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Example Question #354 : Calculus Ii

Find $\lim_{x\rightarrow 5}\frac{{(x-5)}^5}{sin(x-5)}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

$\lim_{x\rightarrow 5}\frac{{(x-5})^5}{sin(x-5)}$

By L'Hopital's Rule:

$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$ if $\frac{f(x)}{g(x)}=\frac{0}{0}$

Since this limit gives us $\frac{0}{0}$ we take the derivative of the numerator and denominator:

$\lim_{x\rightarrow 5}\frac{{(x-5})^5}{sin(x-5)}=\lim_{x\rightarrow 5}\frac{{5(x-5})^4}{cos(x-5)}=0$

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Example Question #355 : Calculus Ii

Find $\lim_{x\rightarrow \frac{\pi}{2}^{-}}tan(x)$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

We know that $tan(\frac{\pi}{2})$ is undefined. However, an easier way to determine this limit is to break up tangent into component parts:

$\lim_{x\rightarrow \frac{\pi}{2}^{-}}tan(x)=\lim_{x\rightarrow \frac{\pi}{2}^{-}}\frac{sin(x)}{cos(x)}$

Since it's approaching $\frac{\pi}{2}$ from the left, cos(x) approaches from positive values and sin(x) approaches from positive values. Since that's the case,

$\lim_{x\rightarrow \frac{\pi}{2}^{-}}tan(x)=\lim_{x\rightarrow \frac{\pi}{2}^{-}}\frac{sin(x)}{cos(x)}=\frac{1}{0}=\infty$

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Example Question #354 : Limits

$\lim_{x\rightarrow -3} \frac{x^2-9}{x+3}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

To subsitute the value into the function, we must first factor the function to remove the hole.

x^2-9=(x-3)(x+3) , using difference of squares.

Therefore, $\frac{(x-3)(x+3)}{(x+3)}=x-3$

Now, we can just substitute! -3-3=-6.

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Example Question #311 : Finding Limits And One Sided Limits

$\lim{x\rightarrow \infty} \; \frac{3x^3+2x^2+x}{(x+2)(x-2)}$

Possible Answers:

$-\infty$

$\infty$

DNE

Correct answer:

$\infty$

Explanation:

Whenever dealing with infinity limits, we only need to consider the highest power of the numinator and denominator. Therefore, we need to foil the denominator first.

$\frac{3x^3+2x^2+x}{x^2-4}$ .

If we only consider the highest powers (everything else will be insignificant when plugging in huge numbers), we only have $\frac{3x^3}{x^2}=3x$ . If you plug infinity into this, you will be get out infinity.

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Example Question #355 : Limits

$\lim_{x \rightarrow 2}\frac{x^2-4}{x+2}$

Possible Answers:

DNE

Correct answer:

Explanation:

This question is the best case scenario for limits. Since the denominator does not equal zero when we plug in the value, we can just substitute the value into the function!

$\frac{2^2-4}{(2+2)}=\frac{0}{4}=0$ .

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Example Question #355 : Calculus Ii

Evaluating the following right sided limit:

$\lim_{x\rightarrow 1^{+}}{\frac{x}{lnx}}$

Possible Answers:

$-\infty$

Undefined

$\infty$

Correct answer:

$\infty$

Explanation:

To evaulate this one sided limit, it is first worht noting that the function is not defined at x=1, due to the fact that ln(1)=0 causing division by zero. Furthermore, the two-sided limit is not defined since x is decreasing for the interval (0,1) and (1,e). This tells us that from the right side when x becomes arbitarly close to 1 but less than e, the function bcomes increasingly large. Also, the function is not defined for all x<0, since ln(x) is undefined for all x<0. This allows us to conclude that for values of x greater than 1 but less than e the function is tending twoards infinity.

Alternatively, for values of x on the interval (1,e), ln(x) is between (0,1), therefore 1/lnx or x/lnx will become increasingly large.

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Example Question #356 : Limits

Find the limit

$\lim_{x \to \infty}e^{\frac{3}{x}}$

Possible Answers:

DNE

Correct answer:

Explanation:

Substituting $x=\infty$

we find that

$e^{\frac{3}{\infty}}\approx e^0=1$

As such

$\lim_{x \to \infty}e^{\frac{3}{x}}=1$

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #351 : Limits

Example Question #352 : Limits

Example Question #353 : Limits

Example Question #354 : Calculus Ii

Example Question #355 : Calculus Ii

Example Question #354 : Limits

Example Question #311 : Finding Limits And One Sided Limits

Example Question #355 : Limits

Example Question #355 : Calculus Ii

Example Question #356 : Limits

All Calculus 2 Resources

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