Pre-Calculus - Determine if a Function is Continuous Using Limits

Find the domain where the following function is continuous:

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

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

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

$(-\infty ,\infty)$

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

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

Explanation

The function in the numerator factors to:

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.

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

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

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

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

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

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

Explanation

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

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

There is no discontinuity at all at . The function simply evaluates to zero at this point.

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

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.

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

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

Explanation

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

Taking the limit from the lefthand side of the function towards a specific point exists.
Taking the limit from the righthand side of the function towards a specific point exists.
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,

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

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

Determine if is continuous on all points of its domain.

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 limit does not match the righthand limit.

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

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

Explanation

First, find that at any point where , .

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.

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

$(-\infty, \infty)$

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

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

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

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

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 , we get: