While studying limits , I came across the following identity :
Let $\lim_{x \to a} f(x) = l$ and $ \lim_{x \to a} g(x) = m$. If $l$ and $m$ exist , then
$\lim_{x \to a}(f+g)(x) = \lim_{x \to a}f(x) +\lim_{x \to a}g(x)$
My question is , is the above identity valid if and only if both $f(x)$ and $g(x)$ are indeterminate individually ?
For example , recently I tried a question ,
If $f(x) = \frac{2}{x-3} , g(x) = \frac{x-3}{x+4}$ and $h(x)= -\frac{2(2x+1)}{x^2+x-12}$ then find the value of
$ \lim_{x \to 3} [f(x) + g(x) + h(x)]$
As we can clearly see that $g(x)$ is not indeterminate for the given limit and hence not all of the functions are indeterminate . Hence when I tried to split the limit and solve , I was unable to get the correct answer. However after taking LCM the whole thing becomes indeterminate and using L’Hospital’s Rule I was able to arrive at the correct answer. So is the rule imperative that all functions must be of indeterminate form ?
The identity is valid
that is in all the cases which don't lead to an indeterminate form $\infty-\infty$.
Note that you can't apply the result to the limit
$$\lim_{x \to 3} [f(x) + g(x) + h(x)]$$
since the single limits for $f(x)$ and $h(x)$ do not exist for $x \to 3$.
And we cannot apply neither for one side limit since
$$\lim_{x \to 3^+} [f(x)]=+\infty \qquad \lim_{x \to 3^-} [f(x)]=-\infty$$
but
$$\lim_{x \to 3^+} [h(x)]=-\infty \qquad \lim_{x \to 3^-} [h(x)]=+\infty$$