What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?

Question

What is the limit of $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?

I attempted the problem via L^Hopital's Rule so I rewrote it as

$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$

then took the natural of both sides

$$\ln(y)=\ln(\frac{\cos(x)-1+\frac{x^2}{2}}{x^4})$$

then using the properties of logarithms I came to the conclusion that

$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+\ln(x^4)$$ $$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x)$$

So then I took the limit as the $\ln(y)$ approaches 0.

$$\lim_\limits {x \to 0} (\ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x))$$

Here I used L'Hopital's Rule and got

$$\lim_\limits {x \to 0} \frac{-\sin(x)+x}{\cos(x)-1+x^2}+\frac{4}{x}$$

then got a common denominator

$$\lim_\limits {x \to 0} \frac{-\sin(x)+x+4 \cdot (\cos(x)-1+x^2)}{x\cdot (\cos(x)-1+x^2)}$$

I clearly made a mistake somewhere because the denominator is 0. The answer by the way is $\frac{1}{24}$. I have no idea how to arrive at that conclusion.

Answer 1

Using Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ So $$\cos(x)-1+\frac{x^2}{2}=\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}=\frac{1}{24}-\frac{x^2}{720}+O\left(x^4\right)$$ which shows the limit and how it is approached.

Answer 2

I believe the problem should read $$\lim_{x \to 0} \frac{\cos(x) - 1 + \tfrac {x^2}2}{x^4};$$ i.e., the sign of the $1$ should be flipped. Otherwise the top goes to $2$ and the bottom to $0$ so the limit is $+\infty$. As I have written it, you can simply use l'Hopital's rule several times: \begin{align*}\lim_{x \to 0} \frac{\cos(x) - 1 + \tfrac {x^2}2}{x^4} &= \lim_{x \to 0} \frac{-\sin(x) +x }{4x^3} \\ &= \lim_{x \to 0} \frac{-\cos(x) + 1}{12x^2} \\ &= \lim_{x\to 0} \frac{\sin(x)}{24x}\\ &= \lim_{x\to 0}\frac{\cos(x)}{24} = \frac 1 {24}.\end{align*}

Answer 3

Here is another take with minimum number of applications of L'Hospital's Rule. We have \begin{align} L &= \lim_{x \to 0}\dfrac{\cos x - 1 + \dfrac{x^{2}}{2}}{x^{4}}\notag\\ &= \lim_{t \to 0}\dfrac{\cos 2t - 1 + 2t^{2}}{16t^{4}}\text{ (putting }x = 2t)\notag\\ &= \frac{1}{16}\lim_{t \to 0}\dfrac{2t^{2} - 2\sin^{2}t}{t^{4}}\notag\\ &= \frac{1}{8}\lim_{t \to 0}\dfrac{t^{2} - \sin^{2}t}{t^{4}}\notag\\ &= \frac{1}{8}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\cdot\frac{t + \sin t}{t}\notag\\ &= \frac{1}{8}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\cdot\left(1 + \frac{\sin t}{t}\right)\notag\\ &= \frac{1}{4}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\notag\\ &= \frac{1}{4}\lim_{t \to 0}\dfrac{1 - \cos t}{3t^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{12}\lim_{t \to 0}\dfrac{1 - \cos^{2} t}{t^{2}(1 + \cos t)}\notag\\ &= \frac{1}{24}\lim_{t \to 0}\dfrac{\sin^{2} t}{t^{2}}\notag\\ &= \frac{1}{24}\notag \end{align} If you have studied Taylor series expansions then that is the best technique to solve such problems (see the answer by Claude Leibovici for details).