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Precalculus : Limits

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Example Questions

Example Question #14 : Find The Limit Of A Function

Find the limit of the function:

$\lim_{x\rightarrow 5^{+}} \frac{5}{5-x}$

Possible Answers:

$\frac{1}{5}$

The limit can't be determined.

$-\infty$

$\infty$

Correct answer:

$-\infty$

Explanation:

Substituting the value of x=5 will yield $\frac{5}{0}$ , which is not in indeterminate form. Therefore, L'Hopital's rule cannot be used.

The question asks for the limit as x approaches to five from the right side in the graph. As the graph approaches closer and closer to x=5 , the y-value decreases to negative infinity and will never touch x=5 .

The correct answer is: $-\infty$

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Example Question #21 : Limits

Find the limit.

$\lim_{x\rightarrow 3}\frac{2x-6}{x-3}$

Possible Answers:

=12

Does not exist

Correct answer:

Explanation:

In the unsimplified form, the limit does not exist; however, the numerator can be factored and simplified.

$\lim_{x\rightarrow 3}\frac{2x-6}{x-3}$

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

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Example Question #11 : Find The Limit Of A Function

Find the limit: $\lim_{x \to 6} \frac{x-3}{3}$

Possible Answers:

$\textup{Undefined}$

Correct answer:

Explanation:

The first and only step is to substitute the value of into the function. Since there is not a zero denominator or an indeterminate form, we do not have to worry about L'Hopital or an undefined limit.

$\lim_{x \to 6} \frac{x-3}{3} = \frac{6-3}{3} = \frac{3}{3} = 1$

The limit will approach to .

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Example Question #22 : Limits

Find the following limit:

$\lim_{x\rightarrow \infty} \frac{3x^2+1}{2x^2}$

Possible Answers:

$\infty$

undefined

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

To solve, simply realize you are dealing with a limit whose numerator and demominator have the same max power. Thus the limit is simply the division of their coefficient.

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

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Example Question #1 : Determine If A Function Is Continuous Using Limits

Find the domain where the following function is continuous:

$\frac{x^4-x^3-12x^2}{x+3}$

Possible Answers:

$(-\infty ,\infty)$

$(-\infty ,4),(4,\infty)$

$(-\infty ,-3],[-3,\infty)$

$(-\infty ,-3),(-3,\infty)$

$(-\infty ,3),(3,\infty)$

Correct answer:

$(-\infty ,-3),(-3,\infty)$

Explanation:

The function in the numerator factors to:

(x-4)(x+3)x^2

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The $\frac{x+3}{x+3}$ gives us a hole at x=-3 so our function is not continuous at x=-3.

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Example Question #1 : Determine If A Function Is Continuous Using Limits

What are the discontinuities in the following function and what are their types?

$\frac{(x+1)(x-3)}{(x-3)(x-4)}$

Possible Answers:

$\textup{Removable at}\ x=3\textup{; Infinite at}\ x=4$

$\textup{Infinite at}\ x=3\ \textup{and}\ x=4$

$\textup{Removable at}\ x=1\ \textup{and}\ x=3$

$\textup{Removable at}\ x=3\textup{; Infinite at}\ x=1$

Correct answer:

$\textup{Removable at}\ x=3\textup{; Infinite at}\ x=4$

Explanation:

Since the factor (x-3) is in the numerator and the denominator, there is a removable discontinuity at x=3 . The function is not defined at x=3 , but function would move towards the same point for the resultant function $\frac{x+1}{x-4}$ .

Since the factor (x-4) cannot be factored out, there is an infinite discontinuity at x=4 . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at x=-1 . The function simply evaluates to zero at this point.

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Example Question #662 : Pre Calculus

Determine if the function $y=x^\frac{2}{3}$ is continuous at x=0 using limits.

Possible Answers:

No, it is not continuous because the lefthand and righthand limits do not equal the value of the function at 0.

Yes, it is continuous because the lefthand and righthand limits are equal.

Yes, it is continuous because the righthand and lefthand limits are equal to the actual value of the function.

No, it is not continuous because the lefthand limit does not match the righthand limit.

Correct answer:

Yes, it is continuous because the righthand and lefthand limits are equal to the actual value of the function.

Explanation:

In order to determine if a function is continuous at a point three things must happen.

1) Taking the limit from the lefthand side of the function towards a specific point exists.

2) Taking the limit from the righthand side of the function towards a specific point exists.

3) The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

1) $\lim_{x->0^{-}}=0$

2) $\lim_{x->0^{+}}=0$

3) y(0)=0

Because all of these conditions are met, the function is continuous at 0.

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Example Question #1 : Determine If A Function Is Continuous Using Limits

Determine if y=e^x is continuous on all points of its domain.

Possible Answers:

Yes, it is continuous because the righthand and lefthand limits are equal to the actual values of the function.

No, it is not continuous because the lefthand and righthand limits do not equal the value of the function at 0.

No, it is not continuous because the lefthand limit does not match the righthand limit.

Yes, it is continuous because the lefthand and righthand limits are equal.

Correct answer:

Yes, it is continuous because the righthand and lefthand limits are equal to the actual values of the function.

Explanation:

First, find that at any point where x= b , y=e^b .

Then find that

$\lim_{x->b^{+}} e^{x}=e^{b}$ and $\lim_{x->b^{-}} e^{x}=e^{b}$ .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

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Example Question #5 : Determine If A Function Is Continuous Using Limits

Let $f(x)=\frac{x^{2}-9}{x-3}$ . Determine if the function is continuous using limits.

Possible Answers:

$\left(-\infty, -\frac{1}{3}\right) \left(-\frac{1}{3},\infty \right)$

$(-\infty, 3) (3,\infty )$

$(-\infty, -3) (-3,\infty )$

$\left(-\infty, \frac{1}{3}\right) \left(\frac{1}{3},\infty \right)$

$(-\infty, \infty)$

Correct answer:

$(-\infty, \infty)$

Explanation:

As approaches , the function $f(x)=\frac{x^{2}-9}{x-3}$ approaches $\frac{3^{2}-9}{3-3}=\frac{0}{0}$ , which is undefined. However, if we factor f(x) , we get:

$f(x)=\frac{x^{2}-9}{x-3}$

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

The (x-3) factors in the numerator and denominator cancel out, leaving f(x)=x+3 .

Therefore, our function is continuous at all values of from $(-\infty, \infty)$ .

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Precalculus : Limits

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #14 : Find The Limit Of A Function

Example Question #21 : Limits

Example Question #11 : Find The Limit Of A Function

Example Question #22 : Limits

Example Question #1 : Determine If A Function Is Continuous Using Limits

Example Question #1 : Determine If A Function Is Continuous Using Limits

Example Question #662 : Pre Calculus

Example Question #1 : Determine If A Function Is Continuous Using Limits

Example Question #5 : Determine If A Function Is Continuous Using Limits

All Precalculus Resources

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