How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$.

Question

Can someone help me with this limit? I'm working on it for hours and cant figure it out.

$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$

I started transforming to the form $ \lim_{x\to 0} e^{ {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}} }$ and applied the l'Hopital rule (since indeterminated $\frac00$), getting:

$$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$

From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?

Answer 1

If we are allowed to use Series Expansion,

$\tan x=x+\dfrac{x^3}3+O(x^5)$

$$\lim_{x\to0}\left(\dfrac{\tan x}x\right)^{1/x^2}=\left(\lim_{x\to0}\left(1+\dfrac{x^2}3+O(x^4)\right)^{1/(1+x^2/3+O(x^4))}\right)^{\lim_{x\to0}\dfrac{1+x^2/3+O(x^4)}{x^2}}$$

Set $\dfrac1{1+x^2/3+O(x^4)}=n$ in the inner limit

Can you take it from here!

Answer 2

It is in the standard form of $1^{\infty}$

The general value of these type of limits is $e^{\lim_{ x \rightarrow 0}{(f(x)-1)g(x)}}$

Here $f(x)=\frac{\tan x}{x},g(x)=\frac{1}{x^2}$

Therefore you need to find $\lim_{x \rightarrow 0} \frac{\tan x-x}{x^3}$

Now, apply L Hospital to get $\lim_{x \rightarrow 0} \frac{ \tan x -x}{x^3}=\frac{ \tan ^2 x}{3x^2}=\frac{1}{3}$

Thus the limit is $e^{\frac{1}{3}}$

Answer 3

If $\lim_{x\rightarrow a}f(x)=1$ and $\lim_{x\rightarrow a}g(x)=\infty$, then, $$\lim_{x\rightarrow a}f(x)^{g(x)}=e^{\lim_{x\rightarrow a}(f(x)-1)g(x)}$$

This is because: Let $f$ and $g$ be functions such that $\lim_{x\rightarrow a}f=1$ and $\lim_{x\rightarrow a}g=\infty$.

Let $$L=\lim_{x\rightarrow a}f^g$$ $$\log L=\log(\lim_{x\rightarrow a}f^g)$$ $$\log L=\lim_{x\rightarrow a}\log(f^g)$$ $$\log L=\lim_{x\rightarrow a}g\log(f)$$ $$L=e^{\lim_{x\rightarrow a}g\log(f)}$$ $$L=e^{\lim_{x\rightarrow a}g\frac{\log(f)}{f-1}(f-1)}$$ $$L=e^{\lim_{x\rightarrow a}g\lim_{x\rightarrow a}\frac{\log(f)}{f-1}\lim_{x\rightarrow a}(f-1)}$$ $$\lim_{x\rightarrow a}\frac{\log(f)}{f-1}=1$$ Thus, $$L=e^{\lim_{x\rightarrow a}g(f-1)}$$

Here, $$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}=\lim_{x\to 0} e^{\left(\frac {\tan x }{x} -1\right){\frac{1}{x^2}}}=\lim_{x\to 0} e^{\left(\frac {\tan x -x}{x^3}\right)}=\lim_{x\to 0} e^{\left(\frac {x+\frac{x^3}{3}+O(x^5) -x}{x^3}\right)}=e^{\frac13}$$

Answer 4

In the same spirit as other answers, consider $$A=\left(\frac {\tan (x) }{x} \right)^{\frac{1}{x^2}}$$ Take logarithms $$\log(A)=\frac{1}{x^2}\log\left(\frac {\tan (x) }{x} \right)$$ Now, consider Taylor series $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\frac {\tan (x) }{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)$$ $$\log\left(\frac {\tan (x) }{x} \right)=\log\left(1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)\right)$$ Now use $\log(1+y)=y-\frac{1}{2}y^2+O\left(y^3\right)$ and replace $y$ by $\frac{x^2}{3}+\frac{2 x^4}{15}$ to get $$\log(A)=\frac{1}{x^2} \left(\frac{x^2}{3}+\frac{7 x^4}{90}+O\left(x^5\right)\right)=\frac{1}{3}+\frac{7 x^2}{90}+O\left(x^3\right)$$ which shows the limit and also how it is approached.

You can continue using $A=e^{\log(A)}$ and still using Taylor arrive to $$A=\sqrt[3]{e}+\frac{7}{90} \sqrt[3]{e} x^2+O\left(x^3\right)$$

Answer 5

Let $\displaystyle y(x) =\ln\left[\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}\right] =\frac{1}{x^2}\ln\left(\frac{\tan x}{x}\right)$, then by using the L'Hôpital's rule several times, \begin{align} \lim_{x\to 0}y(x)&=\lim_{x\to 0}\frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\\ &\stackrel{{\rm H}}{=} \lim_{x\to 0}\frac{\frac{x}{\tan x}\cdot\frac{x\sec^2x-\tan x}{x^2}}{2x}\tag{1}\\ &=\lim_{x\to 0}\frac{x\sec^2x-\tan x}{2x^2\tan x}\\ &\stackrel{{\rm H}}{=} \lim_{x\to 0}\frac{\sec^2x+2x\sec^2x\tan x-\sec^2 x}{4x\tan x+2x^2\sec^2x}\\ &=\lim_{x\to 0}\frac{2\sec^2x\tan x}{4\tan x+2x\sec^2x}\\ &=\lim_{x\to 0}\frac{2\sec^2x}{4+2\left(\frac{x}{\sin x}\right)\sec x}\\ &=\frac{2\lim_{x\to 0}\sec^2x}{4+2\left(\lim_{x\to 0}\frac{x}{\sin x}\right)\left(\lim_{x\to 0}\sec x\right)}\\ &=\frac{1}{3}. \end{align} Hence $\displaystyle \lim_{x\to0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} =\lim_{x\to0}e^{y(x)} =e^{\lim_{x\to0}y(x)}=e^{\frac{1}{3}}$. Notice that $(1)$ holds because $$\lim_{x\to0}\ln\left(\frac{\tan x}{x}\right) =\ln\left(\lim_{x\to 0}\frac{\tan x}{x}\right) \stackrel{{\rm H}}{=} \ln\left(\lim_{x\to 0}\frac{\sec^2 x}{1}\right)=\ln(1)=0,$$ which follows that $\displaystyle \frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\rightarrow\frac{0}{0}$ and we can use the L'Hôpital's rule.

Answer 6

Whenever we have an expression where both base and exponent are variables, it is best to take logs. Thus if $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{1/x^{2}}\right\}\notag\\ &= \lim_{x \to 0}\log\left(\frac{\tan x}{x}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{\tan x}{x}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{\tan x - x}{x}\cdot\dfrac{\log\left(1 + \dfrac{\tan x - x}{x}\right)}{\dfrac{\tan x - x}{x}}\notag\\ &= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot 1\notag\\ &= \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\ &= \frac{1}{3}\notag \end{align} Hence $L = e^{1/3}$.

Answer 7

$$\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac1{x^2}} =\lim_{x\to 0}\exp\left(\frac{1}{x^2}\ln\left(\frac{\tan x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{1}{3}\frac{\ln\left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}\right)= \color{blue}{\exp(\frac13)}$$

Given that $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$