Finding out a limit $\lim_{x\rightarrow 0} \frac{\cos x - e^{- \frac{x^2}{2}}}{x^4}$

Question

I have trouble finding $$\large \lim_{x\rightarrow 0} \frac{\cos x - e^{- \frac{x^2}{2}}}{x^4}.$$ I can use L'Hospital's rule, but it seems slightly ineffective there. Can you help me, please?

Answer 1

Recall that, as $x \to 0$, we have $$ \begin{align} \cos x &=1-\frac{x^2}{2}+\frac{x^4}{24}+\mathcal{O}(x^6)\\\\ e^{-\frac{x^2}{2}}&=1-\frac{x^2}{2}+\frac{x^4}{8}+\mathcal{O}(x^6) \end{align} $$ then $$ \frac{\cos x - e^{- \frac{x^2}{2}}}{x^4}=\frac{\frac{x^4}{24}-\frac{x^4}{8}+\mathcal{O}(x^6)}{x^4}=-\frac{1}{12}+\mathcal{O}(x^2) $$ giving the desired limit.

Answer 2

The consecutive derivatives of numerator/denominator are

$$\begin{align}&(1)\;\;\frac{-\sin x+xe^{-x^2/2}}{4x^3}\\{}\\ &(2)\;\;\frac{-\cos x+e^{-x^2/2}-x^2e^{-x^2/2}}{12x^2}\\{}\\ &(3)\;\;\frac{\sin x-3xe^{-x^2/2}+x^3e^{-x^2/2}}{24x}\\{}\\ &(4)\;\;\frac{\cos x-3e^{-x^2/2}+6x^2e^{-x^2/2}-x^4e^{-x^2/2}}{24}\xrightarrow[x\to 0]{}\frac{1-3}{24}=-\frac1{12}\end{align}$$

Not that terrible, in fact.

Answer 3

You have to apply L'Hopitals rule four times to get a result that way. It is probably quicker to use a series expansion on the numerator. You will need to take the series to at least the term in $x^4$ to get the result.

Answer 4

Use Taylor at order $4$:

• $\cos x=1-\dfrac{x^2}2+\dfrac{x^4}{24}+o(x^4)$
• $\mathrm e^{-\tfrac{x^2}2}=1- \dfrac{x^2}2+\dfrac{x^4}{8}+o(x^4)$

Hence $$\frac{\cos x - e^{- \frac{x^2}{2}}}{x^4}=\frac{\cfrac{x^4}{24}-\dfrac{x^4}{8}+o(x^4)}{x^4}=\frac1{24}-\frac{1}{8}+o(1)=-\frac{1}{12}+o(1).$$

Thus $\,\lim_{x\to 0} \dfrac{\cos x - e^{- \tfrac{x^2}{2}}}{x^4}=-\dfrac 1{12}.$

Answer 5

This solution uses the following standard limits only: \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}} &=&\frac{1}{24} \\ && \\ \lim_{u\rightarrow 0}\frac{1-u-e^{-u}}{u^{2}} &=&-\frac{1}{2}. \end{eqnarray*} Re-write the original expression as \begin{equation*} \frac{\cos x-e^{-x^{2}/2}}{x^{4}}=\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{% x^{4}}\right) +\frac{1}{\left( \sqrt{2}\right) ^{4}}\cdot \left( \frac{% 1-\left( \frac{x}{\sqrt{2}}\right) ^{2}-e^{-\left( \frac{x}{\sqrt{2}}\right) ^{2}}}{\left( \left( \frac{x}{\sqrt{2}}\right) ^{2}\right) ^{2}}\right) \end{equation*} Since \begin{equation*} \lim_{x\rightarrow 0}\left( \frac{x}{\sqrt{2}}\right) ^{2}=0, \end{equation*} then, substitute $u=\left( \frac{x}{\sqrt{2}}\right) ^{2}$ in the second limit, one gets \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\cos x-e^{-x^{2}/2}}{x^{4}} &=&\lim_{x\rightarrow 0}\frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}+\frac{1}{\left( \sqrt{2}\right) ^{4}}\cdot \lim_{u\rightarrow 0}\left( \frac{1-u-e^{-u}}{u^{2}}\right) \\ &=&\frac{1}{24}+\frac{1}{\left( \sqrt{2}\right) ^{4}}\left( -\frac{1}{2}% \right) \\ &=&-\frac{1}{12}. \end{eqnarray*}