mathop {lim }limits_{x to 0} ,left( {frac{{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \left( {\frac{{x - \sin x}}{x}} ight)\sin \left( {\frac{1}{x}} ight)$.We can rewrite the expression

Vedclass · Accepted Answer

equals $0$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \left( {\frac{{x - \sin x}}{x}} ight)\sin \left( {\frac{1}{x}} ight)$.We can rewrite the expression as:$L = \mathop {\lim }\limits_{x 	o 0} \left( {1 - \frac{\sin x}{x}} ight) \sin \left( {\frac{1}{x}} ight)$.We know that $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {\lim }\limits_{x 	o 0} \left( {1 - \frac{\sin x}{x}} ight) = 1 - 1 = 0$.The term $\sin \left( {\frac{1}{x}} ight)$ oscillates between $-1$ and $1$ as $x 	o 0$,meaning it is a bounded function.By the Squeeze Theorem,since the limit of the first part is $0$ and the second part is bounded,the product is $0 	imes (	ext{bounded value}) = 0$.Therefore,the limit is $0$.

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

Similar Questions

The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

If $[.]$ denotes the greatest integer function,then $\lim _{x \rightarrow 0} x^7 \left[ \frac{1}{x^3} \right]$ is equal to

$\lim _{x \rightarrow 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]=$

$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right)$ is equal to

The value of $\lim _{x \rightarrow 0^{+}} \frac{x}{p} \left[ \frac{q}{x} \right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module