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Find $$\lim_{x\to0}\left(x^2\sin\frac1x\right)$$

Solution:

We have:

$$\color{red}{-1\le\sin\frac1x\le1}\;;\;x\in(-5,5)-\{0\}\quad\text{Why?}$$

(Multiplying by $x^2$ …Note that $x^2$ is non-negative)

$$\implies-x^2\le x^2\sin\frac1x\le x^2\;;\;x\in(-5,5)-\{0\}$$

We have

$$\lim_{x\to0}\left(-x^2\right)=0=\lim_{x\to0}\left(x^2\right)$$

$\implies$ By the squeeze theorem,

$$\lim_{x\to0}\left(x^2\sin\frac1x\right)=0$$

Can someone please explain this problem? I could solve the ones without $\sin$/$\cos$.

Why do we assume that $\sin\dfrac1x$ is between $-1$ and $1$?

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    $\begingroup$ sin(x) is always between -1 and 1 ... $\endgroup$ – Hippalectryon Mar 3 '15 at 17:56
  • $\begingroup$ Do you not know that $\sin$ function is bounded by 1? $\endgroup$ – user160738 Mar 3 '15 at 17:56
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    $\begingroup$ Because $-1\leq\sin\alpha\leq1$ is true for any $\alpha$. $\endgroup$ – drhab Mar 3 '15 at 17:57
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    $\begingroup$ Because -1 <= sin(y) <=1 for all Real y (the sine function only takes values between -1 and 1). Let y = 1/x $\endgroup$ – Peter Webb Mar 3 '15 at 17:59
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$ \sin(x)$ or $ \sin(1/x) $ or $ \sin (f(x)) $ is bounded within limits $ \pm 1. $

So x^2 |$ \pm $ 1| tends to 0 as x $ \Rightarrow $ 0.

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