Evaluate $\lim_{x\to 0} \frac{5^x-3^x}{x}$

Question

Problem: Evaluate

$$\lim_{x\to 0} \dfrac{5^x-3^x}{x}$$

This would become quite easy if I could apply L'Hopital's Rule (the answer I got was $\ln(5)-\ln(3)$). However, I'm supposed to solve it without L'Hopital's Rule.
I thought of taking the logarithm of the Numerator and the Denominator, but was unable to proceed further.
Any help on this question would be truly appreciated. Many thanks!$$$$ Cheers!

Answer 1

HINT : Let $f(x)=5^x-3^x$. Then, we have $$\lim_{x\to 0}\frac{5^x-3^x}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}.$$

Answer 2

Here's how you do it using standard limit.

$$L=\lim_\limits{x\to 0}\frac{(5^x-1)-(3^x-1)}{x}$$

$$L=\lim_\limits{x\to 0}\frac{(5^x-1)}{x}-\frac{(3^x-1)}{x}$$

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Lemma:

$$P=\lim_\limits{x\to 0}\frac{a^x-1}{x}=\ln a$$

Put $a^x-1=t$ so that $x=\log_a{(1+t)}$ and notice that as $x\to 0$$\implies$ $a \to 0$

$$P=\lim_\limits{a\to 0}\frac{t}{\log_a(1+t)}=\lim_\limits{a\to 0}\frac{1}{\log_a(1+t)^{1/t}}=\frac{1}{\log_ae}=\ln a$$

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Using this result, the required limit .

$L=\ln5-\ln 3$

Answer 3

If you know

$${\mathrm e}^{k\,x}=1 + k\,x+{\mathcal{O}}(x^2),$$

you can use

$$a^x={\mathrm e}^{x\log(a)}$$

to see that

$$\dfrac{a^x-b^x}{x}=\log(a)-\log(b)+{\mathcal{O}}(x).$$