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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.

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  • $\begingroup$ If that's really what Mathematica says, and you didn't make a typing mistake, then Mathematica is wrong: doing l'Hospital the limit is $\;\infty\;$ . $\endgroup$ – Timbuc Nov 5 '14 at 3:36
  • $\begingroup$ This was a typo, the limit should now be correct. $\endgroup$ – Zach Halle Nov 5 '14 at 3:40
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$$ \lim_{x \rightarrow 0} \frac{(1-\cos x)^2}{\log (1 + \sin^4x)} = \lim_{x \rightarrow 0} \frac{(2\sin^2( x/2))^2)}{\log (1 + \sin^4x)}=1\cdot\lim_{x \rightarrow 0} \frac{4\sin^4(x/2))}{\sin^4x} $$ 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 $\displaystyle\lim_{y\to0}\dfrac{\ln(1+y)}{y}=1$.

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  • $\begingroup$ Oh jeez, this was a typo on my part. The square term should be applied to the whole numerator. I'll edit my question to reflect this. $\endgroup$ – Zach Halle Nov 5 '14 at 3:40
  • $\begingroup$ Who downvoted an old version of answer (to old version of question)?! Now it is corrected to be an answer to new version of question. $\endgroup$ – Przemysław Scherwentke Nov 5 '14 at 3:50
  • $\begingroup$ Some people really rushed to downvote... @Prz , would you mind to explain your second equality in the first line? $\endgroup$ – Timbuc Nov 5 '14 at 3:50
  • $\begingroup$ @Timbuc Enlarged. $\endgroup$ – Przemysław Scherwentke Nov 5 '14 at 3:57
  • $\begingroup$ @PrzemysławScherwentke, so in fact you're using an approximation: $\;\log(1+\sin^4x)\sim\sin^4x\;$ for values close to zero, right? But, even if this approx. is done without Taylor polynomials/series (which require derivatives), how do you prove $\;\frac{\log(1+y)}y\xrightarrow[y\to 0^+]{}1\;$ without l'Hospital or something similar? $\endgroup$ – Timbuc Nov 5 '14 at 4:03
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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 $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$

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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.

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  • $\begingroup$ nice indeed..:) $\endgroup$ – saudade May 12 '15 at 7:10

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