Question

lim [x^2 sin(1/x)/(sinx) =1

x to 0

True or false with full explanation

Answer 1

$L=\displaystyle \lim_{x\to 0} \frac{x^2 sin(\frac{1}{x})}{sin \ x}=\displaystyle \lim_{x\to 0} \frac{x sin(\frac{1}{x})}{\frac{sin \ x}{x}}$ $=$ $\displaystyle \lim_{x\to 0} \frac{\frac{sin(1/x)}{1/x}}{\frac{sin \ x}{x}}$

$=$ $\frac{\displaystyle \lim_{x\to 0}{\frac{sin(1/x)}{1/x}}}{\displaystyle \lim_{x\to 0}\frac{sin \ x}{x}}$

Now, $\displaystyle \lim_{x\to 0}\frac{sin \ x}{x}=1$

So , $L=\displaystyle \lim_{x\to 0}{\frac{sin(1/x)}{1/x}}=\displaystyle \lim_{x\to 0} x\ sin (1/x)$

and $\displaystyle \lim_{x\to 0} x\ sin\ \theta =0 \ \forall \ \theta \isin \ R$

$L=\displaystyle \lim_{x\to 0} x\ sin (1/x)=0$

Hence, the statement $L=\displaystyle \lim_{x\to 0} \frac{x^2 sin(\frac{1}{x})}{sin \ x}=1$ is wrong.