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Find $$\lim_{x\rightarrow 0}\frac{(4^x-1)^3}{\sin\left(\frac{x}{a}\right)\log\left(1+\frac{x^2}{3}\right)}$$

I initially changed $4^x$ to $e^{x \ln(x)}$ and later tried to manipulate the function using L'Hopital's rule. But then I concluded that that was not enough for evaluation; more simplification is required.

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  • $\begingroup$ I think L'Hopital's is your best bet here. $\endgroup$ – SalmonKiller Jun 5 '15 at 14:29
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Use Taylor series.

As $x \rightarrow 0$:

$$4^x = 1 + x \ln 4 + o(x)$$

$$\sin x = x + o(x)$$ $$\ln(1 + x) = x + o(x)$$

So:

$$\lim_{x\rightarrow 0} \dfrac{(4^x-1)^3}{\sin\left(\dfrac{x}{a}\right)\ln\left(1+\dfrac{x^2}{3}\right)} = \lim_{x\rightarrow 0}\dfrac{(x \ln4)^3}{\left(\frac{x}{a}\right)\left(\frac{x^2}{3}\right)} = 3 a (\ln 4)^3$$

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  • $\begingroup$ Use of asymptotics are often superior to use of L'Hospital's Rule! Well done. $\endgroup$ – Mark Viola Jun 5 '15 at 14:38
  • $\begingroup$ why you don't put $O(x)$ I mean Big-Oh $\endgroup$ – Lucas Jun 5 '15 at 15:28
  • $\begingroup$ @Lucas $o(x)$ is stronger than $O(x)$. Also, this is the standard Peano form of the remainder, see en.wikipedia.org/wiki/… $\endgroup$ – lisyarus Jun 5 '15 at 16:00
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And now time for without Taylor or lhopital as anticipated.

$$L=\lim_{x\rightarrow 0} \dfrac{3 \times a\times \dfrac{(4^x-1)^3}{x^3}}{\dfrac{\sin\left(\dfrac{x}{a}\right)}{\dfrac{x}{a}}\times \dfrac{\log\left(1+\dfrac{x^2}{3}\right)}{\dfrac{x^2}{3}}}$$

Notice how i have written the expression here and it is same as the expression you have mentioned.

We will now use three standard limits given by.

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

2)$\lim_\limits{u\to 0}\frac{\sin u}{u}=1$

3)$\lim_\limits{v\to 0}\frac{\ln (1+v)}{v}=1$

Using these limits we can get.

$L=3a \times (\ln 4)^3$

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  • $\begingroup$ Thank you for 3) $\lim_\limits{v\to 0}\frac{\ln (1+v)}{v}=1$ If you can tell more about this point, how one can get this result.. $\endgroup$ – Prateek Jun 5 '15 at 17:01
  • $\begingroup$ @Prateek. It is a quick consequence of L'Hospital. But you can also prove it without l'Hospital using some inequalities involving logarithm. $\endgroup$ – Idris Jun 5 '15 at 18:30
  • $\begingroup$ It's the definition of $\ln e$ you could say, the same can be rewritten as, $\ln (1+v)^{1/v}$ the inner limit is clearly the definition of e. @Prateek ,btw you may like to join here, chat.stackexchange.com/rooms/23760/… mostly indian but group kinda froze atm. ^^ $\endgroup$ – Mann Jun 6 '15 at 5:31

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