Limit of form $ \frac{a^{x} - 1} {x} $ as $ x \to 0 $

Question

I have this question where the part that I cannot understand what to do and how to start the problem. The answer options are given in the form of ln. But, the only logarithmic property of limits I know so far is: $$ \lim_{x \to 0} \frac{\ln(1+x)} {x} = 1 $$

But in this case, the full question is: $$ \lim_{x\to 0} \frac{3 ^{x} - 5^{x} }{x} $$

So, is it just a matter of some manipulation of the previous known property, or is there some special theory required for me to cover? I am actually new to calculus and we didn't cover the L' Hopital's rule and other things, so a solution or hint without using that would be appreciated! Thanks in advance.

Answer 1

You know that $$\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$$ the proof of which is here.

So, for your question, $$\lim_{x\to 0}\frac{3^x-5^x}x=\lim_{x\to 0}\frac{(3^x-1)-(5^x-1)}{x}=\ln 3-\ln 5$$ This completes the answer.

Answer 2

we can do the following:

Take $t=a^x-1 \longrightarrow x=\log_{a}(t+1)=(\log{a})^{-1}\log(t+1)$.

And note that $t\to0$ as $x\to0$

Then, we want to calculate

$$\log{a}\cdot\lim_{t\to0}\frac{t}{log(t+1)}$$ Which you already now that is equal to $\log{a}\cdot1$ but can be calculate by using Taylor expansion is you don't want to use Lhopital.