Evaluate $\lim_{x \rightarrow 0} \frac{(1-\cos x)^2}{\log (1 + \sin^4x)} $

Question

According to Mathematica,

$$\displaystyle\lim_{x \rightarrow 0} \frac{(1-\cos x)^2}{\log (1 + \sin^4x)} = \frac{1}{4}$$

For my purposes it is sufficient to know this limit exists and is finite, but I may not use L'Hopital's rule (or anything relying on differentiability). However, I am stuck on how to show this limit holds. I've tried giving an epsilon-delta proof, but it seems needlessly complicated. Ideally I should be able to show the limit exists using only trig rules, algebra, and eventually reducing this out of being an indeterminate form so that I can simply plug in $0$.

However the problem with this is that I can't seem to break the log term. It seems like I would need to factor its argument, but to do so I need to manipulate $1 + \sin^4 x$ so that I can factor it and separate out the terms. But nothing has worked thus far.

Answer 1

$$ \lim_{x \rightarrow 0} \frac{(1-\cos x)^2}{\log (1 + \sin^4x)} = \lim_{x \rightarrow 0} \frac{(2\sin^2( x/2))^2)}{\log (1 + \sin^4x)}=1\cdot\lim_{x \rightarrow 0} \frac{4\sin^4(x/2))}{\sin^4x} $$ and this behaves as $$ \frac{4x^4}{2^4x^4}, $$ which indeed tends to $1/4$.

We used $1-\cos x=(\sin^2(x/2)+\cos^2(x/2))-(\cos^2(x/2)-\sin^2(x/2))$ and $\displaystyle\lim_{y\to0}\dfrac{\ln(1+y)}{y}=1$.

Answer 2

HINT:

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

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

$$=\frac1{\dfrac{\ln(1+\sin^4x)}{\sin^4x}}\cdot\frac1{(1+\cos x)^2}$$

Now $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$

Answer 3

I always rember that $1-\cos x \sim \frac{x^2}{2}$ and $\ln(1+x) \sim x$ as $x\to 0$. When I need to compute a limit, I always use it.