Find:
L=lim
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?
Answer
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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