Question

If $f\left(x\right)=\frac{{\tan }^{-1}(x+2)}{|x+2|},x\ne -2$ and if $f(-2)=-1$ then at $x=-2$, $f$ is

• A. left continuous
• B. right continuous
• C. continuous
• D. none

Answer 1

The correct option is A left continuous

We have

$f\left(x\right)=\frac{{\tan }^{-1}\left(x+2\right)}{\left|x+2\right|},x\ne -2$

And,

$f(-2)=-1$

Now

$\underset{x\to -2}{lim}\frac{{\tan }^{-1}\left(x+2\right)}{-\left(x+2\right)}=-1$

$\Rightarrow Leftcontinuousatx=-2$

Hence the option $\left(A\right)$ is the correct answer.