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

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 ?

• Welcome to math.se. I recommend visiting this page for a primer on how to type equations using $\LaTeX$ and MathJax to make your equations more readable. Jun 6 '15 at 17:45
• thanks for leting me know. was looking for something like that. Jun 6 '15 at 17:49
• Your question title provides no indication of the question. You should change it Jun 6 '15 at 22:09

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 $$\lim_{x\to0}\frac{1-\cos x}{x}=0, \lim_{x\to0}\frac{x}{1-\cos x}=\infty, \lim_{x\to0}(1+x)^{\frac1x}=e,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.$$

$$\frac{1-\cos x}x=\frac x2+o(x)$$ 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 $\,\mathrm e^{1/2}$.

• what 'o' means ? at o(x) ? how you got to $$\frac{x}{2} + o(x) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaqaaiaadIhaaeaacaaIYaaaaiabgUcaRiaad+gapaGaaiik % a8qacaWG4bWdaiaacMcaaaa!3C39!$$ Jun 7 '15 at 7:01
• @idandi: it is the notion for rest of the terms in the Taylor polynomial Jun 7 '15 at 9:41
• It is Landau notation: roughly speaking, if $f(x)$ and $g(x)$ are two functions in a neighbourhood of $0$, and if $\,g(x)\ne 0$ (except at $x=0$), we say $f(x)=_0o(g(x)\,$ if $\,\lim\limits_{x\to0}\dfrac{f(x)}{g(x)}=0$. Here, intuitively, it means $o(x)$ tends to $0$ much faster than $x$. Jun 7 '15 at 9:55

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*}

• this looks so life changing , but i didnt learn this yet. only next week. Jun 6 '15 at 17:59
• I used a rule called L'Hopital's rule. if you have a quotient/fraction and taking the limit gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then you can differentiate the numerator and the denominator and then take the limit. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
– user230715
Jun 6 '15 at 18:03
• @George learning to use \left and \right will improve the parentheses in this answer, making it mor readable. Jun 6 '15 at 23:32

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

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

• Nice (for the sake of simplicity, the first term expansion is enough here for the log, I believe -- this will make things easier in the proof).\ Jun 6 '15 at 17:51
• what 'o' means ? at o(x) ? i know log identities. just dont know what this 'o' means Jun 6 '15 at 18:09
• $f=o(g)\iff \lim\frac{f}{g}=0$ Jun 6 '15 at 18:11
• (actually, this is not exactly that as $g$ could techncially have zeros. It means there exists a function $\varepsilon$ going to $0$ such that $f=\varepsilon\cdot g$; but the definition @marwalix gives above is good enough for most purposes, and easier to parse.) Jun 6 '15 at 18:52