Chapter 14.3, Problem 72AYU

To determine

To find: The numbers at which $f$ is continuous.

Answer to Problem 72AYU

Continuous for all real values of $x$ excluding $x=3$ and $x\le 0$.

Interval of continuity: $\left(0,\infty \right)-\left\{3\right\}$.

Explanation of Solution

Given:

$f\left(x\right)=\frac{\text{ln}x}{x-3}$

Calculation:

$f\left(x\right)=\frac{\text{ln}x}{x-3}$

The numerator function $\text{ln}x$ is defined only for $x>0$.

So $f\left(x\right)$ is undefined for $x\le 0$.

Moreover, the denominator function becomes zero when $\left(x-3\right)=0$; ie $x=3$.

So $f\left(x\right)$ is undefined for $x=3$.

Since $f\left(x\right)=\frac{\text{ln}x}{x-3}$ is undefined for $x\le 0,x=3$; therefore $f\left(x\right)=\frac{\text{ln}x}{x-3}$ is continuous for all real numbers excluding $x\le 0$ and $x=3$.

Thus the interval on which $f$ is continuous is $\left(0,\infty \right)-\left\{3\right\}$.