If f(x) = begin{cases} x e^{-left( frac{1}{|x | vedclass

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

Question

(B) Given $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} ight)}, & x 
e 0 \ 0, & x = 0 \end{cases}$.For continuity at $x = 0$:$R.H

Vedclass · Accepted Answer

Continuous for all $x$ but not differentiable at $x = 0$

Vedclass · Answer

(B) Given $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} ight)}, & x 
e 0 \ 0, & x = 0 \end{cases}$.For continuity at $x = 0$:$R.H.L. = \lim_{h 	o 0^+} f(0+h) = \lim_{h 	o 0^+} h e^{-\left( \frac{1}{h} + \frac{1}{h} ight)} = \lim_{h 	o 0^+} h e^{-2/h} = 0$.$L.H.L. = \lim_{h 	o 0^+} f(0-h) = \lim_{h 	o 0^+} (-h) e^{-\left( \frac{1}{h} - \frac{1}{h} ight)} = \lim_{h 	o 0^+} (-h) e^0 = 0$.Since $R.H.L. = L.H.L. = f(0) = 0$,$f(x)$ is continuous at $x = 0$.For differentiability at $x = 0$:$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h e^{-2/h} - 0}{h} = \lim_{h 	o 0^+} e^{-2/h} = 0$.$Lf'(0) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{-h e^0 - 0}{-h} = 1$.Since $Rf'(0) 
e Lf'(0)$,$f(x)$ is not differentiable at $x = 0$.

If $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

Similar Questions

In the interval $(-2 \pi, 0)$, the function $f(x) = \sin \left(\frac{1}{x^3}\right)$

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} a^2 \cos^2 x + b^2 \sin^2 x, & x \leq 0 \\ e^{ax+b}, & x > 0 \end{cases}$ and is a continuous function,then:

If $\{x\} = x - [x]$ where $[x]$ is the greatest integer $\leq x$ and $\lim_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^4} = \theta$,then $\tan \theta =$

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...

On the interval $I = [-2, 2]$,the function $f(x) = \begin{cases} (x + 1) e^{-\left[ \frac{1}{|x|} + \frac{1}{x} \right]} & x \neq 0 \\ 0 & x = 0 \end{cases}$ is given. Which one of the following does not hold good?

Vedclass Test Series

Exam Paper Generator

Online Exam Module