Answer
The statement is false.
Work Step by Step
This statement is false. Here's an example to disprove:
$\lim\limits_{x\to5}f(x)=\lim\limits_{x\to5}(x^2-5x)=5^2-5\times5=0$
$\lim\limits_{x\to5}g(x)=\lim\limits_{x\to5}(x-5)=5-5=0$
However, $$\lim\limits_{x\to5}\frac{x^2-5x}{x-5}$$$$=\lim\limits_{x\to5}\frac{x(x-5)}{x-5}$$$$=\lim\limits_{x\to5}x$$$$=5$$
That means $\lim\limits_{x\to5}\frac{x^2-5x}{x-5}$ does exist.
