One more Limit

Since we know that $\displaystyle \lim_{x \to 0} \dfrac{x}{\sin x}=1$ , therefore \[ \begin{array}{} \displaystyle \lim_{x\to0} \left( \dfrac{ x^2 \sin \frac1x } {\sin x} \right) & = \displaystyle \lim_{x\to0} \left( x \cdot \sin \frac1x \right) \\ & = \displaystyle \lim_{x\to 0} \left( x \cdot \{ \text{Some value belonging to [-1,1]} \} \right) \\ & = 0. \square \end{array} \]

Let $u=\frac{1}{x}$ as $x\to 0\,\,,\,\,u\to \infty$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}\sin \frac{1}{x}}{\sin x}=\underset{u\to \infty }{\mathop{\lim }}\,\frac{\frac{1}{{{u}^{2}}}\sin u}{\sin \frac{1}{u}}=\underset{u\to \infty }{\mathop{\lim }}\,\frac{\frac{1}{u}\frac{\sin u}{u}}{\sin \frac{1}{u}}=\underset{u\to \infty }{\mathop{\lim }}\,\frac{\sin u}{u}\times \underset{u\to \infty }{\mathop{\lim }}\,\frac{\frac{1}{u}}{\sin \frac{1}{u}}$

We know that , $-1\le \sin u\le 1\Leftrightarrow \frac{-1}{u}\le \frac{1}{u}\sin u\le \frac{1}{u}\Leftrightarrow \underset{u\to \infty }{\mathop{\lim }}\,\frac{-1}{u}\le \underset{u\to \infty }{\mathop{\lim }}\,\frac{\sin u}{u}\le \underset{u\to \infty }{\mathop{\lim }}\,\frac{1}{u}$

As $u\to \infty \,\,\,,\,\,\frac{1}{u}\to 0$ so by sandwish theorem $\underset{u\to \infty }{\mathop{\lim }}\,\frac{\sin u}{u}=0$

Also $\underset{u\to \infty }{\mathop{\lim }}\,\frac{\frac{1}{u}}{\sin \frac{1}{u}}=\underset{\frac{1}{u}\to 0}{\mathop{\lim }}\,\frac{\frac{1}{u}}{\sin \frac{1}{u}}=\underset{w\to 0}{\mathop{\lim }}\,\frac{w-0}{\sin w-\sin \left( 0 \right)}=\frac{1}{\frac{d}{dw}{{\left( \sin w \right)}_{w=0}}}=\frac{1}{\cos 0}=1$

Hence $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}\sin \frac{1}{x}}{\sin x}=1\times 0=0$

Sin(1/x)/sin(x) = "something"

So, it's the limit for x^2 × "something"

Substitute 0 in for x and you end up multiplying "something" by zero.

Works for me.

consider this solution : (x/sinx)*(sin(1/x)/(1/x))=1

Isn't this supposed to be 1?