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As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...

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Note that by Taylor's series

  • $\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4)$
  • $\sin x^2=x^2+o(x^2)$

thus $$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-x^2+\frac{x^4}{2}+o(x^4)}{x^4+o(x^4)}=\frac{\frac12+o(1)}{1+o(1)}\to \frac12$$

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Note that

$${x^2-\log(1+x^2)\over x^2\sin^2x}={x^2\over\sin^2x}\cdot{x^2-\log(1+x^2)\over x^4}={x^2\over\sin^2x}\cdot{u-\log(1+u)\over u^2}$$

where $u=x^2\to0^+$ as $x\to0$. If we take ${x\over\sin x}\to1$ for granted, then L'Hopital takes care of the rest:

$$\lim_{u\to0^+}{u-\log(1+u)\over u^2}=\lim_{u\to0}{1-{1\over1+u}\over2u}=\lim_{u\to0}{1\over2(1+u)}={1\over2}$$

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  • $\begingroup$ I've just posted the same solution 10 sec before you :) (+1) for affinity $\endgroup$ – user Feb 5 '18 at 23:58
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    $\begingroup$ @gimusi, I just noticed. It looks like you were 13 seconds quicker! Oh, and it looks like Prasun Biswas got there before either of us! $\endgroup$ – Barry Cipra Feb 5 '18 at 23:58
  • $\begingroup$ Yes I can see now! $\endgroup$ – user Feb 6 '18 at 0:02
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$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{\dfrac 1{x^2}-\dfrac{\log(1+x^2)}{x^4}}{\dfrac{\sin^2x}{x^2}}$$

Taking $z=x^2$, note that the numerator becomes $$\dfrac 1z-\dfrac{\log(1+z)}{z^2}=\dfrac{z-\log(1+z)}{z^2}\to\dfrac{1-\frac 1{1+z}}{2z}\to\dfrac{1}{2(z+1)^2}\to\frac 12$$

as $z=x^2\to 0$ as $x\to 0$ where the final arrowed steps use L'Hopital

The denominator goes to $1$ since $\dfrac{\sin x}x\to 1$ as $x\to 0$

Thus, the limit is $1/2$ by a final application of L'Hopital's rule.

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  • $\begingroup$ If you simplify the ${1-{1\over1+z}\over2z}$ first, you don't need the second L'Hopitation. $\endgroup$ – Barry Cipra Feb 6 '18 at 0:02
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    $\begingroup$ Then it is slightly different form our solution :) $\endgroup$ – user Feb 6 '18 at 0:03
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As an alternative

$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-\log(1+x^2)}{x^4}\cdot\frac{x^2}{\sin^2x}\to \frac12$$

indeed

$\frac{x^2}{\sin^2x}\to 1$ by standard limit

and let $y=x^2\to 0$

$$\frac{x^2-\log(1+x^2)}{x^4}=\frac{y-\log(1+y)}{y^2}\stackrel{HR}\implies\frac{1-\frac1{1+y}}{2y}=\frac{1+y-1}{2y(1+y)}=\frac{1}{2(1+y)}\to\frac12$$

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\begin{align*} \lim_{x\rightarrow 0}\dfrac{x^{2}-\log(1+x^{2})}{x^{2}\sin^{2}x}&=\lim_{x\rightarrow 0}\dfrac{x^{2}-x^{2}+\dfrac{1}{2}x^{4}-\dfrac{1}{3}x^{6}\cdots}{x^{2}\sin^{2}x}\\ &=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}-\dfrac{1}{3}x^{4}\cdots}{\sin^{2}x}\\ &=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)}{\sin^{2}x}\\ &=\lim_{x\rightarrow 0}\left(\dfrac{1}{2}\cdot\dfrac{x^{2}}{\sin^{2}x}\right)\cdot\lim_{x\rightarrow 0}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)\\ &=\dfrac{1}{2}. \end{align*}

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$$\lim_\limits{x\to0}\frac{x^2-x^2+\frac{x^4}{2}+\mathcal{O}\left(x^5\right)}{x^4+\mathcal{O}\left(x^5\right)}=\lim_\limits{x\to0}\frac12\cdot\frac{x^4+\mathcal{O}\left(x^5\right)}{x^4+\mathcal{O}\left(x^5\right)}=\frac12.$$

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