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  • If <img alt="f(x)=\left\{\begin{matrix} \frac{e^{[x]+\left | x \right |}-2}{[x]+\left | x \right |}, &x\neq 0\\ -1, & x=0 \end{matrix}\right." src="https://learn.careers360.com/latex-image/

If f(x)=\left\{\begin{matrix} \frac{e^{[x]+\left | x \right |}-2}{[x]+\left | x \right |}, &x\neq 0\\ -1, & x=0 \end{matrix}\right. (where [\cdot ] denotes G.I.F) then

Option: 1

f(x) is continuous at x=0
 


Option: 2

\lim_{x\rightarrow 0^{+}}f(x)=-1

 


Option: 3

\lim_{x\rightarrow 0^{-}}f(x)=1


Option: 4

none of these


Answers (1)

best_answer

 

Continuity at a point -

A function f(x)  is said to be continuous at  x = a in its domain if 

1.  f(a) is defined  : at  x = a.

2. \lim_{x\rightarrow a}\:f(x)\:exists\:means\:limit\:x\rightarrow a

of  f(x) at  x = a exists from left and right.

3. \lim_{x\rightarrow a}\:f(x)=f(a)\:then\:the\:limit\:equals \:the\:value\:at\:x=a

-

 

 

\because \:\:L.H.L. =\lim_{x\rightarrow 0^{-}}f(x)= \lim_{x\rightarrow 0^{-}}\frac{e^{\left [ x \right ]+\left | x \right |}-2}{\left [ x \right ]+\left | x \right |}=\frac{e^{-1}-2}{-1}

and \because \:\:R.H.L. =\lim_{x\rightarrow 0^{+}}f(x)= \lim_{x\rightarrow 0^{+}}\frac{e^{\left [ x \right ]+\left | x \right |}-2}{\left [ x \right ]+\left | x \right |}

=\lim_{x\rightarrow 0^{+}}\frac{e^{x}-2}{x}\rightarrow -\infty \:\:\:\:\therefore L.H.L.\neq R.H.L.

\therefore (D)

Posted by

Ritika Kankaria

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