<s>[INST] 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?
 [/INST] or, for $A > 1,$
$$ A^x = e^{x \log A} = 1 + x \log A + O( x^2)   $$ so you have
$$ \frac{1 + x \log 5 -1 - x \log 3 + O(x^2)}{1 + x \log 3 -1 - x \log 2 + O(x^2)} = \frac{x (\log 5 - \log 3) + O(x^2)}{x (\log 3 - \log 2) + O(x^2)} = \frac{ \log 5 - \log 3 + O(x)}{ \log 3 - \log 2 + O(x)} $$ 
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