limit

Question

Value of $\lim_{x \to 0} \frac{x^2 \sin \left(\frac{1}{x}\right)} {\sin x}$ is

Comments

should be 0

---

please explain the steps ...

---

Ans is 0.

Explanation

lim x->0 x² sin(1/x)/sin x

lim x->0 sin(1/x)/(1/x)/sinx/x

lim x->0 sin(1/x)/(1/x)/lim x->0 sinx/x

let for numerator ie lim x->0 sin(1/x)/1/x we say y=1/x and it can written as lim x->infinity sinx/x ie =0

and for denominator ie lim x->0 sinx/x we know is 1 ie 0/1=0 Ans.

Please correct me if I am wrong.

---

Some standard results which are useful to answer this question:-

$1)$ $\large{\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)}=0$ 

                Answer for the above result.

$2)$ $\large{\lim_{x \to 0}\dfrac{x}{\sin x}} = 1 $ ( this is just an inversion of standard limit $\lim_{x \to 0} \dfrac{\sin x }{x} = 1$.)

$ $

Coming to the question we have to evaluate,   $\large{\lim_{x \to 0}\dfrac{x^2\sin(\frac{1}{x})}{\sin x}}$

= $\large{\lim_{x \to 0}\dfrac{x^2\sin(\frac{1}{x})}{\sin x} = \lim_{x \to 0 }\dfrac{x}{\sin x} (x \sin(\frac{1}{x}))} = (1)(0) = 0$.

so $\large{\lim_{x \to 0}\dfrac{x^2\sin(\frac{1}{x})}{\sin x}} = 0$

Answer 1

$\lim_{x \to 0} \frac{x^2 \sin \left(\frac{1}{x}\right)} {\sin x} \\=\frac{ \lim_{x \to 0} \frac{\sin \left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)}} {\lim_{x \to 0} \frac{ \sin x}{x}} \\=\frac{\lim_{ y \to \infty } \frac{\sin y} {y}}{1} \\= 0 (\because \sin y \leq 1, y \to \infty) .$

Comments

Arjun sir, i think here we have to consider two cases while using substitution because if i say,  $\large{y = \dfrac{1}{x}}$,

Case1):- $ \large{x \to 0^-,y \to -\infty}$

Case2):- $\large{x \to 0^+, y\to +\infty}$

For the case 2 answer is same as above, coming to case 1:-

we have to evaluate,     $\large{\lim_{y \to -\infty}\dfrac{\sin y}{y}}$

we knew that, $|\sin x| \leq 1$

$= \dfrac{|\sin x |}{|x|} \leq \dfrac{1}{|x|}$

As $\lim_{y \to -\infty} \dfrac{1}{|x|} = 0$ so, $\lim_{y \to -\infty}\left|\dfrac{\sin x}{x}\right| = 0$

in both case we the limit value of the final answer will be 0.