Evaluating: $\lim\limits_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$
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:
- 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 $\dfrac{a-1}{1/b}$.
- 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
$$\log{\left(\frac{\sin{x}}{x}\right)} \sim \log{\left(1-\frac{x^2}{6}\right)} \sim -\frac{x^2}{6}$$
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:
$$\Large e^{g(x)(f(x)-1)}$$
PROOF:
$$\Large \lim_{x\to a} f(x)^{g(x)}\\$$ $$\Large = \lim_{x\to a} e^{\ln f(x)^{g(x)}}$$ $$\Large = \lim_{x\to a} e^{g(x).\ln f(x)}$$ (...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:
$${\Large e^\left({\lim\limits_{x\to 0}6\dfrac{\sin x-x}{x^3}}\right)}$$