Any ideas on how to tackle this limit? $$\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$$
I tried many ways but only got more complex stages, not easier ones...
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take the logarithm – user153330 Jan 31 '15 at 12:52
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did you try it? it doesnt work – Bak1139 Jan 31 '15 at 13:05
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related problem. – Mhenni Benghorbal Jan 31 '15 at 14:52
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Related : https://math.stackexchange.com/questions/1770823/how-to-find-lim-limits-x-to-0-left-frac-tan-x-x-right-frac1x – Arnaud D. Sep 09 '21 at 14:36
4 Answers
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Notice that
$$\tan x=x+\frac{x^3}3+o(x^3)$$ so $$\left(\frac{\tan x}{x}\right)^{1/{x^2}}\sim_0\left[\left(1+\frac{x^2}{3}\right)^{3/x^2}\right]^{1/3}\xrightarrow{x\to0}e^{1/3}\tag{$\because [1+1/x]^x\sim_0e$}$$
RE60K
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1To prove the asymptotic equivalence between $\frac{\tan x}{x}$ and $1+\frac{x^2}{3}$ in a neighbourhood of the origin can be done in many ways, Taylor series are just a tool. The point is that such an asymptotic equivalence trivially gives the value of the limit. – Jack D'Aurizio Jan 31 '15 at 13:04
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I understand that but this is not how the problem is meant to be solved. – Bak1139 Jan 31 '15 at 13:08
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1@Bak1139: then explain which methods are allowed. We do not have clairvoyance. – Jack D'Aurizio Jan 31 '15 at 13:10
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$$\begin{align}\lim_{x \to0 }\left(\frac{\tan x}{x}\right)^{1/x^2} &=\lim_{x \to0 }e^{\displaystyle \left(\frac1{x^2}\ln\frac{\tan x}x\right)} \\&=\lim_{x \to0 }e^{\displaystyle \left(\frac1{x^2}\frac{\ln\left(1+\left(\frac{\tan x}x-1\right)\right)}{\left(\frac{\tan x}x-1\right)}\left(\frac{\tan x}x-1\right)\right)} \\&=\lim_{x\to0}e^{\displaystyle \left(\frac1{x^2}\left(\frac{\tan x}x-1\right)\right)}\tag{$\because \lim_{x\to0}\frac{\ln(1+ x)}x=1$} \\&=\lim_{x\to0}e^{\displaystyle \left(\frac{\tan x-x}{x^3}\right)}\tag{$*$, proved below} \\&=e^{1/3} \end{align}$$ Now to prove $(*)$ we can use L'Hospital: $$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\sec^2x-1}{3x^2}=\lim_{x\to0}\frac{2\sec^2 x\tan x}{6x}=\lim_{x\to0}\frac13\underbrace{\frac{\tan x}{x}}_1\underbrace{\sec^2x}_1=\frac13$$ Or Taylor: $$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{(x+x^3/3+O(x^5))-x}{x^3}=\frac13$$
RE60K
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from stage 1 to two you've lost an x in the denominator...so the continuation is unclear... – Bak1139 Jan 31 '15 at 14:52
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By far the easiest and fastest way: $$\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = \lim_{x \to 0}\exp\left(\frac1{x^2}\ln\left(1 + \frac{x^2}{3}\right)\right) = \lim_{x \to 0}\exp\left(\frac1{x^2}\cdot\frac{x^2}3\right) = e^{1/3}$$
Using the fact that, for $x \to 0$, $$\begin{align} \tan x &\sim x + \frac{x^3}3\\ \ln(1 + x) &\sim x \end{align}$$
rubik
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First, subtracting $(10)$ from $(9)$ in this answer, we get that $$ \lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13\tag{1} $$ Next, since $\lim\limits_{x\to0}\left(1+tx^2\right)^{1/x^2}=e^t$ converges uniformly on compact sets, $$ \begin{align} \lim_{x\to0}\left(\frac{\tan(x)}x\right)^{1/x^2} &=\lim_{x\to0}\left(1+\frac{\tan(x)-x}{x^3}x^2\right)^{1/x^2}\\ &=\lim_{x\to0}\left(1+\frac13x^2\right)^{1/x^2}\\[6pt] &=e^{1/3}\tag{2} \end{align} $$
