# Calculate $\lim_{x\to 0}\frac{ \cos x-e^{\frac{-x^2}{2}}}{x^4}$ [closed]

How do i calculate this limit $$\lim_{x\to 0}\frac{ \cos x-e^{\frac{-x^2}{2}}}{x^4}$$

## closed as off-topic by user223391, Shailesh, user91500, Workaholic, E. JosephNov 25 '16 at 8:28

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If we use the Taylor serie:

\begin{equation} \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4)\\e^{-\frac{x^2}{2}}=1-\frac{x^2}{2}+\frac{x^4}{8}+o(x^4) \end{equation}

we get:

\begin{equation} \lim_{x\to 0}\frac{ \cos x-e^{\frac{-x^2}{2}}}{x^4}=\lim_{x\to 0}\frac{ 1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)-\left(1-\frac{x^2}{2}+\frac{x^4}{8}+o(x^4)\right)}{x^4}=\\=\lim_{x\to 0}\frac{ \frac{x^4}{24}-\frac{x^4}{8}+o(x^4)}{x^4}=\lim_{x\to 0}\frac{-\frac{1}{12}x^4+o(x^4)}{x^4}=-\frac{1}{12} \end{equation}

• I get $-\frac{1}{12}$ also but by a simpler method. – Rene Schipperus Nov 25 '16 at 3:58
• @ReneSchipperus Applying twice de l'Hopital theorem I suppose...:-) – MattG88 Nov 25 '16 at 4:05

This is not an answer but it is too long for a comment.

As shown in answers, Taylor series are a fantastic tool for computing limits. But they can do much more.

In this specific case, continue adding terms for $\cos(x)$ and $e^{-\frac{x^2}{2}}$; you would arrive to $$\frac{\cos (x)-e^{-\frac{x^2}{2}}}{x^4}=-\frac{1}{12}+\frac{7 x^2}{360}-\frac{13 x^4}{5040}+O\left(x^6\right)$$ which, for sure, shows the limit but also how it is approached.

If you plot both functions for $-1 \leq x \leq 1$ on the same graph, you will not be able to see the difference.