Can you show me a good approach for taking the limit of this function?

Question

I tried to use binomial expansion, but I didn't get the same result. I would like to know how to approach this please. I know the answer is $\sqrt{e}$.

My problem is :

$$\lim\limits_{x\to 0} \left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}$$

and also, can I find some good manipulation with this kind of problems ?

Answer 1

You can use this way: \begin{eqnarray} \lim\limits_{x\to 0} \left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}&=&\lim\limits_{x\to 0} \left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{x}{1-\cos x}}\right)^{\frac{1-\cos x}{x^2}}=e^{\frac12} \end{eqnarray} This is because $$ \lim_{x\to0}\frac{1-\cos x}{x}=0, \lim_{x\to0}\frac{x}{1-\cos x}=\infty, \lim_{x\to0}(1+x)^{\frac1x}=e,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12. $$

Answer 2

Hint look at the $\log$ then use $\cos{x}=1-\frac{x^2}{2}+o(x^3)$ and $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$

Let's look in more details

$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{x}\log\left(1+\frac{1-\cos x}{x}\right)$$

Using the expansion of $\cos$ we get

$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{x}\log\left(1+\frac{x}{2}+o(x)\right)$$

And using the expansion of $\log$ we get

$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{2}+o(x)$$

Answer 3

Note that we can write \begin{equation*} \lim_{x\to 0}\exp(\frac{\ln(1+\frac{1-\cos(x)}{x})}{x}). \end{equation*} Applying L'Hopital's rule twice gives us \begin{equation*} \exp(\lim_{x\to 0}\frac{x\cos(x)}{1+2x-\cos(x)+x\sin(x)}) \\ =\exp(\lim_{x\to 0}\frac{x}{1+2x-\cos(x)+x\sin(x)}). \end{equation*} Applying L'Hopital's rule again gives \begin{equation*} \exp(\lim_{x\to 0} \frac{1}{2+x\cos(x)+2\sin(x)})=\sqrt{e}~_{\square} \end{equation*}

Answer 4

$$\frac{1-\cos x}x=\frac x2+o(x)$$ hence $$\ln\Bigl(1+\frac{1-\cos x}x\Bigr)^{\!\tfrac 1x}=\frac1x\ln\Bigl(1+\frac x2+o(x)\Bigr)=\frac1x\Bigl(\frac x2+o(x)\Bigr)=\frac12+o(1)$$ so that the limit is $\,\mathrm e^{1/2}$.