The limit of x sin left(e^{frac{1}{x}}right) | vedclass

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

Question

(A) We need to evaluate $\lim_{x ightarrow 0} x \sin \left(e^{\frac{1}{x}}ight)$.Since $-1 \leq \sin 	heta \leq 1$ for any real $	heta$, we have

Vedclass · Accepted Answer

is equal to $0$

Vedclass · Answer

(A) We need to evaluate $\lim_{x ightarrow 0} x \sin \left(e^{\frac{1}{x}}ight)$.Since $-1 \leq \sin 	heta \leq 1$ for any real $	heta$, we have $-1 \leq \sin \left(e^{\frac{1}{x}}ight) \leq 1$.Multiplying by $x$ (assuming $x > 0$): $-x \leq x \sin \left(e^{\frac{1}{x}}ight) \leq x$.As $x ightarrow 0^+$, both $-x$ and $x$ approach $0$. By the Squeeze Theorem, $\lim_{x ightarrow 0^+} x \sin \left(e^{\frac{1}{x}}ight) = 0$.For $x  0$. As $x ightarrow 0^-$, $h ightarrow 0^+$.The expression becomes $\lim_{h ightarrow 0^+} (-h) \sin \left(e^{-\frac{1}{h}}ight)$.As $h ightarrow 0^+$, $e^{-\frac{1}{h}} ightarrow e^{-\infty} = 0$, so $\sin \left(e^{-\frac{1}{h}}ight) ightarrow \sin(0) = 0$.Thus, $\lim_{h ightarrow 0^+} (-h) \cdot 0 = 0$.Since the left-hand limit and right-hand limit are both $0$, the limit exists and is equal to $0$.

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

Similar Questions

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

If $f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q} \\ -x, & \text{if } x \in \mathbb{Q}^c \end{cases}$,then $\lim_{x \to 0} f(x)$ is

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = $

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

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module