<s>[INST] What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? 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.
 [/INST] 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*}
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