# How do you find the limit of ((x^2 sin (1/x))/sinx) as x approaches 0?

Sep 7, 2016

0

#### Explanation:

${\lim}_{x \to 0} \frac{{x}^{2} \sin \left(\frac{1}{x}\right)}{\sin} x$

We can split this out as follows
$= {\lim}_{x \to 0} \frac{x}{\sin x} \cdot x \cdot \sin \left(\frac{1}{x}\right)$

and we note that the limit of the product is the product of the known limits
$= \textcolor{red}{{\lim}_{x \to 0} \frac{x}{\sin x}} \cdot \textcolor{g r e e n}{{\lim}_{x \to 0} x} \cdot \textcolor{b l u e}{{\lim}_{x \to 0} \sin \left(\frac{1}{x}\right)}$

The red portion is a well known fundamental trig limit and evaluates to 1

Clearly the green portion evaluates to 0

For the blue portion, we note that $\sin \Psi \in \left[- 1 , 1\right] \forall \Psi$. IOW it is always a finite number within these bounds.

the product of these limits is therefore

$1 \cdot 0 \cdot \left[- 1 , 1\right] = 0$

Sep 7, 2016

$\text{The Reqd. Limit=} 0$.

#### Explanation:

The Reqd. Limit$= {\lim}_{x \rightarrow 0} \frac{{x}^{2} \sin \left(\frac{1}{x}\right)}{\sin} x$

=lim_(xrarr0){(x/sinx)(xsin(1/x)}

={lim_(xrarr0)(x/sinx)}{lim_(xrarr0)(xsin(1/x)}

A well-known Standard Form of Limit states$: {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} x\right) = 1$.

:." The Reqd. Limit reduces to "{lim_(xrarr0)(xsin(1/x)}

To determine this limit, we use another well-known Theorem ,

called The Sandwich Theorem , which states :

If, $f , g , h$ are real-valued functions of a real variable such that,

g(x)<=f(x)<=h(x), &, lim_(xrarra)g(x)=lim_(xrarra)h(x)=l, then

${\lim}_{x \rightarrow a} f \left(x\right) = l$

Now, we know that, $\forall x \in \mathbb{R} , - 1 \le \sin \left(\frac{1}{x}\right) \le 1$

Multiplying this inequality by $x > 0 , - x \le x \sin \left(\frac{1}{x}\right) \le x$

Applying the Sandwich Theorem, since $x > 0$, we get,

${\lim}_{x \rightarrow 0 +} - x \le {\lim}_{x \rightarrow 0 +} x \sin \left(\frac{1}{x}\right) \le {\lim}_{x \rightarrow 0 +} x$.

Knowing that the Left-most and the Right-most Limits are each $0$,

${\lim}_{x \rightarrow 0 +} x \sin \left(\frac{1}{x}\right) = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(0 +\right)$

To find ${\lim}_{x \rightarrow 0 -} x \sin \left(\frac{1}{x}\right)$, we let, x=-t, t>0, so that, as x approaches 0-, t approaches 0+, and hence,

lim_(xrarr0-)xsin(1/x)=lim_(trarr0+){(-t)(sin(1/-t)}

=lim_(trarr0+){(-t)(-sin(1/t)}#

$= {\lim}_{t \rightarrow 0 +} t \sin \left(\frac{1}{t}\right)$

$= 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots b y \left(0 +\right)$, i.e.,

${\lim}_{x \rightarrow 0 -} x \sin \left(\frac{1}{x}\right) = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(0 -\right)$

From $\left(0 +\right) \mathmr{and} , \left(0 -\right)$, we conclude that,

$\text{The Reqd. Limit=} 0$.

Enjoy Maths.!