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  • <img alt="f(x)=\left\{\begin{matrix} \frac{e^{e/x-e^{-e/x}}}{e^{1/x}+e^{-1/x}} , &x\neq 0 \\ k ,&x=0 \end{matrix}\right." src="https://learn.careers360.com/latex-image/?f%28x%29%3D%5Cleft%5

f(x)=\left\{\begin{matrix} \frac{e^{e/x-e^{-e/x}}}{e^{1/x}+e^{-1/x}} , &x\neq 0 \\ k ,&x=0 \end{matrix}\right.

 

Option: 1

f is continuous at x=0, when k=0
 


Option: 2

f is not continuous at x=0 for any real k.


Option: 3

\lim_{x\rightarrow 0}f(x) exist infinitely

 


Option: 4

none of these


Answers (1)

best_answer

 

Condition for discontinuity -

1. \:L\neq R

\lim_{x\rightarrow a^{-}}\:f(x)\neq \lim_{x\rightarrow a^{+}}\:f(x)

limit of function at x = a does not exist.

2.\:L=R\neq V

limit exist but not equal to f(x) at  x = a

-

 

 

\lim_{x\rightarrow 0^{+}}\frac{e^{e/x}-e^{-e/x}}{e^{1/x}+e^{-1/x}}=\lim_{x\rightarrow 0^{+}}\frac{e^{\frac{e-1}{ex}}(1-e^{-2e/x})}{(1+e^{-2/x})}=+\infty

\lim_{x\rightarrow 0^{-}}\frac{e^{e/x}-e^{-e/x}}{e^{1/x}+e^{-1/x}}=\lim_{x\rightarrow 0^{-}}\frac{e^{-e/x}(e^{2e/x}-1)}{e^{-1/x}(e^{+2/x}+1)}

\lim_{x\rightarrow 0^{-}}e^{-\left ( \frac{e-1}{x} \right )}\left ( \frac{e^{2e/x}-1}{e^{2/x}+1} \right )=-\infty

limit doesn't exist So f(x) is discoutinous

Posted by

HARSH KANKARIA

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