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

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?

• Did you tag this (limits-without-lhopital) because you want to disallow solutions using L'Hospital's rule? – Martin Sleziak May 4 '16 at 11:04
• Easy trick ------- $$\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac1{x^2}} =\lim_{x\to 0}\exp\left(\frac{1}{x^2}\ln\left(\frac{\tan x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{1}{3}\frac{\ln\left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}\right)= \color{blue}{\exp(\frac13)}$$ Given that $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$ – Guy Fsone Nov 27 '17 at 15:30

Put $$y = \left({\tan(x)\over x}\right)^{1/x^2}.$$ Then $$\log(y) = {\log(\tan(x)) - \log(x)\over x^2}$$ Can you evaluate the RHS with the help of L'hospital?

• No artificial ingredients: all of my logs are natural. – ncmathsadist Mar 13 '15 at 1:57
• Too funny comment about natural logs. – Paramanand Singh Jun 14 '15 at 7:16

L'Hospital's rule is not the alpha and omega of limits computation!

As $\tan x= x+\dfrac{x^3}3+o(x^3)$, $$\frac{\tan x}x=1+\frac{x^2}3+o(x^2),\enspace\text{hence}\enspace \frac 1{x^2}\ln\Bigl(\frac{\tan x}x\Bigr)=\frac 1{x^2}\ln\Bigl(1+\frac{x^2}3+o(x^2)\Bigr)=\frac 13+o(1)\to \frac 13$$ so that $$\lim_{x\to 0}\Bigl(\frac{\tan x}x\Bigr)^{\tfrac 1{x^2}}=\mathrm e^{\frac 13}.$$

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 $L = e^{1/3}$.

• That should do it. +1 – Mark Viola May 4 '16 at 14:09

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.

It is possible to compute this limit only by making use of some basic limits: \begin{eqnarray*} \lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\ \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\ \lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1. \end{eqnarray*} Take the logarithm \begin{eqnarray*} L &=&\ln \left( \frac{\tan x}{x}\right) ^{1/x^{2}} \\ &=&\frac{1}{x^{2}}\ln \frac{\tan x}{x} \\ &=&\frac{1}{x^{2}}\ln \left( 1+\frac{\tan x}{x}-1\right) \\ &=&\frac{\left[ \frac{\tan x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+% \left[ \frac{\tan x}{x}-1\right] \right) }{\left[ \frac{\tan x}{x}-1\right] } \\ &=&\frac{\tan x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{% \tan x}{x}-1 \end{eqnarray*} since \begin{equation*} \lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\tan x}{x}-1)=1-1=0 \end{equation*} then \begin{equation*} \lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{% \ln (1+u)}{u}=1, \end{equation*} and therefore \begin{equation*} \lim_{x\rightarrow 0}L(x)=\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}\cdot \lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\frac{1}{3}\cdot 1=1. \end{equation*} Backward, one take the exponential, and using continuity arguments, obtains \begin{equation*} \lim_{x\rightarrow 0}\left( \frac{\tan x}{x}\right) ^{1/x^{2}}=\lim_{x\rightarrow 0}e^{L(x)}=e^{1/3}. \end{equation*}

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!

• You're too fast for my slow typing. ;-)) – Mark Viola May 4 '16 at 4:28

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 $$A=\sqrt[3]{e}+\frac{7}{90} \sqrt[3]{e} x^2+O\left(x^3\right)$$

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

• Its best not to use such formulas as you mention. Proper approach is to take logs and simplify. See my answer. – Paramanand Singh May 4 '16 at 7:31
• @ParamanandSingh I added an explanation for the formula – GoodDeeds May 4 '16 at 10:05
• The step by step proof which you gave for the formula is correct, but I still consider it is best to actually write the steps for the question rather than inventing formulas for each and every type of limit problem. This adds to the already existing load of almost infinitely many formulas in calculus. It is best to stick to fundamental rules (algebra of limits, squeeze theorem, standard limits etc) to solve such problems. – Paramanand Singh May 4 '16 at 10:08

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 $e^{\frac{1}{3}}$

• This is not correct. The term is $$e^{\frac{1}{x^2}\left(\log(\tan(x))-\log(x)\right)}$$ – Mark Viola May 4 '16 at 4:32
• Indeed it is correct see here wolframalpha.com/input/… – Legend Killer May 4 '16 at 4:51
• The answer is correct. The method is highly flawed. $f(x)$ is not $\tan(x)/x$. I wrote the correct form in the previous comment. – Mark Viola May 4 '16 at 5:23
• Fully agree with comment from @Dr.MV. The proper technique here is to take logs. See my answer. Such formulas as the one you mention are useless and seem to indicate that calculus is merely a bag of tricks (which it is not). – Paramanand Singh May 4 '16 at 7:29

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

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