<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] 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).
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