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$$\lim_{x\to 0+}\exp\left(\frac{\ln \left( \frac{\ln(1+\tan 4x)}{4x} \right)}{\tan x}\right)=\lim_{x\to 0+}\exp\left(\frac{ \frac{\ln(1+\tan 4x)}{4x} -1}{x}\right)$$

Can anybody explain above equation? I can't understand right hand side of the equation. Where is $\tan x$ and how pop up $-1$? Thank you.

Sorry, infact I solved a question linked my question using Taylor expansion and I'm looking for a solution that not using Taylor and I saw the solution I have mentioned but I can't get it.

deepblue
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  • I cannot understand your edit "Sorry, infact I solved a question linked my question using Taylor expansion and I'm looking for a solution that not using Taylor and I saw the solution I have mentioned but I can't get it." especially the end, from "I saw..." – Anne Bauval Feb 25 '23 at 22:37

1 Answers1

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The notion of asymptotic equivalence is convenient here.

As $x\to0,$

  • $\tan x\sim x$, and
  • $\ln(1+\tan 4x)\sim\tan4x\sim4x$ i.e. $$u:=\frac{\ln(1+\tan4x)}{4x}\to1,\text{ hence }\ln u\sim u-1.$$

Let us now look at the limit of the exponent in your RHS (which will also be the limit of the exponent in your LHS, by the previous observations). $$\frac{u-1}x=\frac{\ln(1+\tan 4x)-4x}{4x^2}=\frac{\tan 4x-\frac{\tan^2(4x)}2+o(x^2)-4x}{4x^2}\sim\frac{-\frac{(4x)^2}2}{4x^2}=-2.$$

So, each of your two limits is equal to $e^{-2}$

Anne Bauval
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