Using L'Hospital's rule doesn't help much as $$ \lim_{x\to 0} \left(1+x^2\right)\left(\frac{\sin\left(\frac 1x\right)\sin^2(x)}{x^2} - \cos\left(\frac 1x\right)\sin(2x)\right) $$ which seems to not exists as $\lim_{x\to 0} \sin(1/x)$ does not exist.

However, using online calculators one can show that the original limit is $0$.

  • $\begingroup$ $cos(1/x)$ is bounded, $\sin(x)$ and $\arctan(x)$ have save series expansion to 1st order around zero. So it will approach 0 linearly. $\gamma \sin(x)^2/\arctan(x) \approx \gamma x^2/x = \gamma x$ $\endgroup$ – Michael Paris Jan 20 at 21:34

Observe that with L'Hospital

$$\lim_{x\to0}\frac{\sin^2x}{\arctan x}=\lim_{x\to0}(1+x^2)\sin2x=1\cdot0=0$$

so the original limit is zero, too, as $\;\cos\frac1x\;$ is bounded

  • 1
    $\begingroup$ Thanks, but why is it OK ignoring $\cos(1/x)$ and applying L'Hopital only to $\sin^2(x)/\arctan(x)$? $\endgroup$ – J.Doe Jan 20 at 21:30
  • $\begingroup$ Note, J.Doe, that $\lim_{x\to 0}\cos(1/x) = \lim){x\to \infty} \cos x$ where $\cos x$ is bounded: $-1 \leq \cos x \leq 1$ for all $x$. $\endgroup$ – jordan_glen Jan 20 at 21:45
  • $\begingroup$ @jordan_glen I intuitively understand it and since I know the answer I can see why this is true, but I want to show it rigorously that I can use L'Hopital only for the non-bounded part of the expression. $\endgroup$ – J.Doe Jan 20 at 21:53
  • $\begingroup$ @J.Doe You have here the limit of an expression that can be split in two parts: for one of them I showed above the limit is zero. The other part is a function bounded in the whole real line and thus also in any neighborhood of zero, of course. Now, the limit of something tending to zero times something bounded is zero. The proof of this fact is very easy, even with the $\;\epsilon-\delta\;$ definition of limit $\endgroup$ – DonAntonio Jan 20 at 21:59

Since $\frac{\sin x}{x}\frac{x}{\arctan x}\to 1$, $\frac{\sin^2 x}{\arctan x}\to 0$. The cosine has modulus $\le 1$, so the limit is $0$.


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