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Find: $$ L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2} $$

My approach:

Because of the fact that the above limit is evaluated as $\frac{0}{0}$, we might want to try the De L' Hospital rule, but that would lead to a more complex limit which is also of the form $\frac{0}{0}$.

What I tried is: $$ L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right) $$ Then, if the limits $$ L_1 = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}}, $$

$$ L_2 = \lim_{x\to0}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right) $$ exist, then $L=L_1L_2$.

For the first one, by making the substitution $u=1-\frac{\sin(x)}{x}$, we have $$ L_1 = \lim_{u\to u_0}\frac{\sin(u)}{u}, $$ where $$ u_0 = \lim_{x\to0}\left(1-\frac{\sin(x)}{x}\right)=0. $$ Consequently, $$ L_1 = \lim_{u\to0}\frac{\sin(u)}{u}=1. $$

Moreover, for the second limit, we apply the De L' Hospital rule twice and we find $L_2=\frac{1}{6}$.

Finally, $L=1\frac{1}{6}=\frac{1}{6}$.

Is this correct?

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    $\begingroup$ The answer is right. The $1/6$ is clear from the power series expansion of $\frac{\sin x}{x}$. $\endgroup$ – André Nicolas Dec 7 '15 at 20:51
  • $\begingroup$ Thank you very much @AndréNicolas! $\endgroup$ – nullgeppetto Dec 7 '15 at 20:52
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    $\begingroup$ You are welcome. That was a nice trick, multiplying and dividing by $1-\frac{\sin x}{x}$. Using L'Hospital's Rule afterwards is OK, but in general when possible I get a better feel of what's happening from the power series than from L'Hospital's Rule. $\endgroup$ – André Nicolas Dec 7 '15 at 20:57
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    $\begingroup$ Better to do $L_2$ as the limit of $\frac{x-\sin x}{x^3}$, which requires L'Hopital three times. You might have to apply some additional knowledge your way. $\endgroup$ – Thomas Andrews Dec 7 '15 at 20:57
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    $\begingroup$ Cleverly done ! $\endgroup$ – Yves Daoust Dec 8 '15 at 11:14
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In a slightly different way, using the Taylor expansion, as $x \to 0$, $$ \sin x=x-\frac{x^3}6+O(x^5) $$ gives $$ 1-\frac{\sin x}x=\frac{x^2}6+O(x^4) $$ then $$ \sin \left( 1-\frac{\sin x}x\right)=\frac{x^2}6+O(x^4) $$ and

$$ \frac{\sin \left( 1-\frac{\sin x}x\right)}{x^2}=\frac16+O(x^2) $$

from which one may conclude easily.

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By L' Hospital anyway:

$$\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$$ yields

$$\cos\left(1-\frac{\sin(x)}x\right)\frac{\sin(x)-x\cos(x)}{2x^3}.$$

The first factor has limit $1$ and can be ignored.

Then with L'Hospital again:

$$\frac{x\sin(x)}{6x^2},$$

which clearly tends to $\dfrac16$.

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    $\begingroup$ Wasn't so horrible after all. $\endgroup$ – Yves Daoust Dec 8 '15 at 11:29

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