Question

solve this problem...​

Answer 1

Hey there!

Given:

$\displaystyle \lim_{x \to 0} \dfrac{\tan(\pi\sin^2x)+(|x|-\sin(x[x]))^2}{x^2}$

Some properties Required for this Question:

$\star \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \ or \ \displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} = 1$

Evaluating the Limit:

$\displaystyle \lim_{x \to 0} \dfrac{\tan(\pi\sin^2x)+(|x|-\sin(x[x]))^2}{x^2}$

Multiplying and dividing by  $\pi(\sin^2x)$,

$\displaystyle \lim_{x \to 0} \dfrac{\tan(\pi\sin^2x)}{x^2} \times \dfrac{\pi\sin^2x}{\pi \sin^2x} + \displaystyle \lim_{x \to 0} \dfrac{(|x|-\sin(x[x]))^2}{x^2}$

Using Properties given above and applying (a-b)²:

$\pi + \displaystyle \lim_{x \to 0} \dfrac{x^2 + \sin^2(x[x]) - 2|x|\sin(x[x])}{x^2}$

$\displaystyle \pi + \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(x[x])}{x^2} - \dfrac{2|x|\sin (x[x])}{x^2}$  

Now, RHL:

$When, \ x \to 0^+ \ then, |x| = x \ and \ [x] = 0$

Substituing in limit:

$\displaystyle \pi + \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(x[x])}{x^2} - \dfrac{2|x|\sin (x[x])}{x^2}$  

$= \pi + 1 + 0 - 0\\\\=\pi + 1$

Now, LHL:

$When, \ x \to 0^-\ then, |x| =- x \ and \ [x] = -1$

Substituting in limit:

$\displaystyle \pi + \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(x[x])}{x^2} - \dfrac{2|x|\sin (x[x])}{x^2}$  

$= \pi + 1 + \displaystyle \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(-x)}{x^2} - \dfrac{2(-x)\sin (-x)}{x^2}$

$= \pi + 1 + \displaystyle \lim_{x \to 0} \dfrac{sin^2x}{x^2} - \dfrac{2\sin x}{x}$

$= \pi + 1 +1 - 2$

$= \pi$

$\because RHL\neq LHL$

Hence, Limit does not exist.

Answer 2

Step-by-step explanation:

Now, RHL:

$</p><p>When, \ x \to 0^+ \ then, |x| = x \ and \ [x] \:$

Substituting in limit:

$</p><p>\displaystyle \pi + \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(x[x])}{x^2} - \dfrac{2|x|\sin (x[x])}{x^2} \:$

$\begin{lgathered}= \pi + 1 + 0 - 0\\\\=\pi + 1\end{lgathered} \:$

Now, LHL:

−

then,∣x∣=−x and [x]=−1

Substituting in limit:

$\displaystyle \pi + \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(x[x])}{x^2} - \dfrac{2|x|\sin (x[x])}{x^2}</p><p>$

$= \pi + 1 + \displaystyle \lim_{x \to 0} \dfrac{x^2}{x^2} + \dfrac{sin^2(-x)}{x^2} - \dfrac{2(-x)\sin (-x)}{x^2}.$

$= \pi + 1 + \displaystyle \lim_{x \to 0} \dfrac{sin^2x}{x^2} - \dfrac{2\sin x}{x}</p><p>.$

$\pi + 1 +1 - 2=.$

$\pi.$

$\because RHL\neq LHl.$

Hence, Limit does not exist.