Find $\lim_{x\rightarrow 0} \frac{5^x-3^x}{3^x-2^x}$ 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?
 A: Yes, it is correct, but it is not correct to say that$$\lim_{x\to0}\frac{5^x\frac{((3/5)^x-1)}{x-0}}{3^x\frac{((2/3)^x-1)}{x-0}}=\lim_{x\to0}\frac{5^xf'(0)}{3^xg'(0)}.$$That is, you can't replace some $x$'s by $0$, while keeping others as $x$.
And you could have used L'Hopital's rule$$\lim_{x\to0}\frac{5^xf(x)}{3^xg(x)}=\lim_{x\to0}\frac{5^x}{3^x}\times\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}=\frac{\log\left(\frac35\right)}{\log\left(\frac23\right)}.$$
A: or, for $A > 1,$
$$ A^x = e^{x \log A} = 1 + x \log A + O( x^2)   $$ so you have
$$ \frac{1 + x \log 5 -1 - x \log 3 + O(x^2)}{1 + x \log 3 -1 - x \log 2 + O(x^2)} = \frac{x (\log 5 - \log 3) + O(x^2)}{x (\log 3 - \log 2) + O(x^2)} = \frac{ \log 5 - \log 3 + O(x)}{ \log 3 - \log 2 + O(x)} $$ 
A: 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 $$\frac{\log 5-\log 3}{\log 3-\log 2}$$
