TITLE: Find $\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$
QUESTION [5 upvotes]: I want to find the limit $$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$$
My efforts:
$$\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}=\lim_{x\rightarrow 0}\frac{5^x((3/5)^x-1)}{3^x((2/3)^x-1)}$$
Multiplying and dividing numerator and denominator by $x$ we get, $$\lim_{x\rightarrow 0}\frac{5^x\frac{((3/5)^x-1)}{x-0}}{3^x\frac{((2/3)^x-1)}{x-0}}\tag{1}$$
Let,
$$f(x)=(3/5)^x-1, g(x)=(2/3)^x-1$$
Now $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=f'(x)$$
and similarly for $g$, we can rewrite $(1)$ as $$\lim_{x\rightarrow 0} \frac{5^x}{3^x}\times \frac{f'(0)}{g'(0)}\tag{2}$$
We know if $h(x)=a^x,$ then $h'(x)=a^x \log(a)$
Computing and putting everything in piece, we write $(2)$ as 
$$\lim_{x\rightarrow 0}\frac{\log(3/5)}{\log(2/3)}$$
So we get limit equal to $\frac{\log(3/5)}{\log(2/3)}$.
Is my computation correct?

REPLY [4 votes]: You may directly use the standard limit $\lim\limits _{x\to 0}\dfrac{a^x-1}{x}=\log a$ and write the expression under limit as $$\dfrac{\dfrac {5^x-1}{x}-\dfrac{3^x-1}{x}}{\dfrac{3^x-1}{x}-\dfrac{2^x-1}{x}} $$ and get the limit as $$\frac{\log 5-\log 3}{\log 3-\log 2}$$