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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!

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  • $\begingroup$ Alright, thanks! $\endgroup$ – Ishan Jun 18 '15 at 12:53
  • $\begingroup$ Please, do not use \dfrac in the titles. And title should not consist only of LaTeX. See: [Guidelines for good use of $\LaTeX$ in question titles(meta.math.stackexchange.com/questions/9687/…) $\endgroup$ – Martin Sleziak Jun 18 '15 at 14:18
  • $\begingroup$ Alright Sir. Sorry for that. $\endgroup$ – Ishan Jun 18 '15 at 14:23
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HINT : Let $f(x)=5^x-3^x$. Then, we have $$\lim_{x\to 0}\frac{5^x-3^x}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}.$$

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  • $\begingroup$ Sorry Sir, I couldn't quite understand . Could you please elaborate a bit more, Sir? $\endgroup$ – Ishan Jun 18 '15 at 12:49
  • $\begingroup$ I just noticed the similarity to $\lim_{x\to 0}\dfrac{x^n-a^n}{x-a}=nx^{n-1}$ $\endgroup$ – Ishan Jun 18 '15 at 12:49
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    $\begingroup$ Note that $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$. $\endgroup$ – mathlove Jun 18 '15 at 12:50
  • $\begingroup$ Alright, thanks Sir. $\endgroup$ – Ishan Jun 18 '15 at 12:53
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    $\begingroup$ IMO, this is just an instance of L'Hospital for $f(x)/x\to f'(x)/1$. $\endgroup$ – Yves Daoust Jun 18 '15 at 14:21
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Here's how you do it using standard limit.

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

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


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$ $a \to 0$

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


Using this result, the required limit .

$L=\ln5-\ln 3$

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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

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

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  • $\begingroup$ Sir, I don't know that $O$ symbol you wrote. Please could you tell me what it is? $\endgroup$ – Ishan Jun 18 '15 at 14:22
  • $\begingroup$ @BetterWorld: My Dear Lord, havet a look at thus linkus. $\endgroup$ – Nikolaj-K Jun 18 '15 at 17:20

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