I am trying to evaluate the following but without result.

$$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$$

Can you please give me some hints? I have tried to put log to both sides but it hasn't lead me somewhere... Thanks a lot


Write it as $$\left[\left(1+\dfrac{\sin(x)-x}{x}\right)^{\dfrac{x}{\sin(x)-x}}\right]^{\dfrac{6(\sin(x)-x)}{x^3}}.$$

Notice what was done is to add $1$ and subtract $1$ in the base of the exponential, and then tweak the exponent, to make it look like $\big(1+\dfrac{1}{n}\big)\large^n$ inside the big brackets.

The part inside the brackets goes to $e$. So, you have to compute the limit of $\dfrac{6(\sin(x)-x)}{x^3}$

General strategy:

This is an indeterminate limit of the form $1^\infty$. You can approach these indeterminate forms in the following two ways:

  1. Take the $a^b$ and write it as $\left[(1+(a-1))^{\dfrac{1}{a-1}}\right]^{\large(a-1)b}$ The part in brackets tends to $e$. So, you only need to resolve the indeterminate form $(a-1)b$. Since this is an indeterminate of the product it could be solved by applying L'Hospital to $\dfrac{a-1}{1/b}$.
  2. Take the $a^b$ and write it as $e^{b\ln(a)}$ and then you only need to resolve the indeterminate form of the product $b\ln(a)$. Which could also be approached by L'Hospital to $\dfrac{\ln(a)}{1/b}$.

The two are essentially the same. The only difference might be that in one you get a logarithm in what remains to be computed and in the other you don't.


Just know that

$$\log{\left(\frac{\sin{x}}{x}\right)} \sim \log{\left(1-\frac{x^2}{6}\right)} \sim -\frac{x^2}{6}$$

  • 3
    $\begingroup$ Nice, it's the point where I am exactly stuck but why is that so? O_o $\endgroup$ – darkchampionz Jul 19 '13 at 12:03
  • $\begingroup$ @darkchampionz: why are you stuck? $\endgroup$ – Ron Gordon Jul 19 '13 at 12:50
  • $\begingroup$ I don't know how to get rid of log(sinx/x) :| $\endgroup$ – darkchampionz Jul 19 '13 at 12:54
  • $\begingroup$ @darkchampionz: That is precisely the point of my answer. The log sinc piece becomes $-x^2/6$. Then when you multiply by $6/x^2$, the log of your limit becomes $-1$. Exponentiate both sides to get your limit. $\endgroup$ – Ron Gordon Jul 19 '13 at 13:06

You have: $$\lim_{x\to0} \bigg(\frac{\sin x}{x}\bigg)^{6/x^2}$$

We know that
$$\lim_{x\to0} \frac{\sin x}{x} = 1$$ (this can be proved by writing the Taylor's Expansion of $\sin x$);

and that $$\frac{6}{x^2} \to\infty$$

So your problem is in the form of $(\to1)^{(\to\infty)}$.
Whenever we reach a situation like the one above, we do the following steps:

If $\lim_{x\to a} f(x)^{g(x)}$, where $f(x)\to1$ and $g(x)\to\infty$, the value of the limit is:

$$\Large e^{g(x)(f(x)-1)}$$


$$\Large \lim_{x\to a} f(x)^{g(x)}\\$$ $$\Large = \lim_{x\to a} e^{\ln f(x)^{g(x)}}$$ $$\Large = \lim_{x\to a} e^{g(x).\ln f(x)}$$ (...using logarithmic properties)

The index part is getting very tiny and squeezed up making it difficult to read - which is why I'm only solving the index below:

\begin{eqnarray} Let L &=& g(x).\ln f(x)\\ &=& g(x).\ln(1 + f(x) - 1)\\ &=& g(x).\frac{\ln\bigg(1 + \big(f(x) - 1\big)\bigg)}{\big(f(x) - 1\big)}.(f(x) - 1)\\ \end{eqnarray} We know that $f(x) - 1 \to0$ $(\because f(x) \to1)$; and that:
$$\lim_{x\to 0} \frac{\ln(1 + x)}{x} = 1$$

So we get: $$L =\lim_{x\to 0} g(x)\cdot(f(x) - 1)$$

Hence, your limit is:

$$\Large e^\left({\lim\limits_{x\to 0} g(x)\cdot(f(x) - 1)}\right)$$

Now, substituting the values for your functions, your limit is:

$${\Large e^\left({\lim\limits_{x\to 0}6\dfrac{\sin x-x}{x^3}}\right)}$$

  • $\begingroup$ Surely for this method to work you need to do if the limit exists in the first place? $\endgroup$ – Andrew D Jul 19 '13 at 13:45
  • $\begingroup$ How can the limit result in functions? $\endgroup$ – Pedro Tamaroff Jul 19 '13 at 13:55
  • $\begingroup$ @PeterTamaroff The question is in the form of $f(x)^{g(x)}$ where $f(x) \to1$ and $g(x) \to\infty$. Since the value of this is $e^{g(x)(f(x)-1)}$, we need to substitute the values; which will ultimately give us $e^{6\frac{(\sin x - x)}{x^3}}.$ Now the limit needs to be solved. $\endgroup$ – mikhailcazi Jul 19 '13 at 14:14
  • $\begingroup$ See, the answer even tallies with @RGB's. $\endgroup$ – mikhailcazi Jul 19 '13 at 14:17
  • $\begingroup$ @AndrewD I don't understand what you mean. :/ $\endgroup$ – mikhailcazi Jul 19 '13 at 14:39

The best hint I can give, I think, is not to give up if an idea doesn't immediately solve the problem for you.

If $$ L = \lim_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$$ then $$ \log L = \lim_{x \to 0} \frac{6 \log\left(\frac{\sin x}{x} \right)}{x^2} $$

Now forget for a moment that you're trying to find $\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$. You have a new problem to see if you can solve:

Evaluate the limit $$\lim_{x \to 0} \frac{6 \log\left(\frac{\sin x}{x} \right)}{x^2} $$

What is the form of this limit? What methods do you have for solving limits of such a form?


Easy trick

$$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac6{x^2}} =\lim_{x\to 0}\exp\left(\frac{6}{x^2}\ln\left(\frac{\sin x -x}{x}+1\right)\right) \color{blue}{\sim \lim_{x\to 0}\exp\left(\frac{\ln\left(1-\frac{x^2}{6}\right)}{\frac{x^2}{6}}\right)= e^{-1}}$$

Given that $$\sin x -x \sim -\frac{x^3}{6}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1-h\right)}{h} = -1$$


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