# The Squeeze Theorem

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 $$1$$?

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

$\sin(x)$ 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.