Question

If $\underset{x\to a}{Lt}[f\left(x\right)+g\left(x\right)]$ exists, then does it imply that $\underset{x\to a}{Lt}f\left(x\right)$ and $\underset{x\to a}{Lt}g\left(x\right)$ also exist?

Answer 1

No, we can show by an example.

Let $f\left(x\right)=x+\frac{1}{x}$

and $g\left(x\right)=x-\frac{1}{x}$

Clearly , f(x) and g(x) are defined for all values of x , except 0.

${lim}_{x\to 0}f\left(x\right)$ does not exists

${lim}_{x\to 0}g\left(x\right)$ does not exists

Now , $f\left(x\right)+g\left(x\right)=x+\frac{1}{x}+x-\frac{1}{x}$

$\Rightarrow f\left(x\right)+g\left(x\right)=2x$

${lim}_{x\to 0}\left[f\right(x)+g(x\left)\right]={lim}_{x\to 0}2x=0$