Question

Evaluate the limit lim x to 0 (sinx[cos (1/x^2) + sin(1/x^2)])

Answer 1

First, we can condense the terms being added as

`\cos\frac1{x^2}+\sin\frac1{x^2}=\sqrt2\sin\left(\frac1{x^2}+\frac\pi4\right)`

because
`\cos x` and
`\sin x` differ in phase by
`\frac\pi4`. Now using the fact that
`|\sin x|\le1`, we have

`-\sin x\le\sin x\left(\cos\frac1{x^2}+\sin\frac1{x^2}\right)\le\sin x`

so that by the squeeze theorem,

`\displaystyle\lim_(x\to0)(-\sin x)\le\lim_(x\to0)\sin x\left(\cos\frac1{x^2}+\sin\frac1{x^2}\right)\le\lim_(x\to0)\sin x`

`\displaystyle0\le\lim_(x\to0)\sin x\left(\cos\frac1{x^2}+\sin\frac1{x^2}\right)\le0`

and so the limit is 0.