Can i use L’Hopital’s rule $f(x)\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{g(x)}}{{h(x)}}} \right]$?

Question

Clarifying L’Hopital’s rule: the form of the function is

$z=\frac{{f(x)g(x)}}{{h(x)}}$.

And $g(0) = h(0) = 0,f(0) \ne 0$.

Can I use L’Hopital rules such as,

$z(0)=f(x)\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{g(x)}}{{h(x)}}} \right]$

instead of $z(0)=\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{f(x)g(x)}}{{h(x)}}} \right]$. Thanks.

Answer 1

L'Hopital's rule say that $\lim_{x \rightarrow 0} \frac{f(x)g(x)}{h(x)} = \lim_{x \rightarrow 0} \frac{(f(x)g(x))'}{h'(x)}$ .

Then you have that $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$.

By a property of limits, $\lim_{x \rightarrow 0} \frac{(f(x)g(x))'}{h'(x)} = \frac{\lim_{x\rightarrow 0} f'(x)g(x)}{\lim_{x \rightarrow 0} h'(x)} + \frac{\lim_{x\rightarrow 0} f(x)g'(x)}{\lim_{x \rightarrow 0} h'(x)}$.

In the first term you have $\lim_{x\rightarrow 0} f'(x)g(x) = f'(a)g(0)$ and this equal zero if you have that $f'(a)$ is finite.

Then you will end with $\lim_{x \rightarrow 0} \frac{f(x)g(x)}{h(x)} = \frac{\lim_{x\rightarrow 0} f(x)g'(x)}{\lim_{x \rightarrow 0} h'(x)} = \lim_{x\rightarrow 0} f(x) \left[ \frac{\lim_{x\rightarrow 0}g'(x)}{\lim_{x \rightarrow 0} h'(x)} \right] $

Answer 2

So you take $z=\frac{{f(x)g(x)}}{{h(x)}}$ and want to use $f(x)\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{g(x)}}{{h(x)}}} \right]$ in place of $\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{f(x)g(x)}}{{h(x)}}} \right]$

Let's consider $g(x)=x,h(x)=x^2$ and $f(x)=x, x\ne 0$ then we have not finite limit in first one, while second limit exists.