The function `f:""R""~""{0}vecR` given by `f(x)=1/x-2/(e^(2x)-1)` can be made continuous at x = 0 by defining f(0) as (1) 2 (2) `-1` (3) 0 (4) 1

Options

A. 2
B. -1
C. 0
D. 1

Answer

D

Explanation

`lim_(xto0)(e^(2x)-1-2x)/(x(e^(2x)-1))`
`=lim_(xto0)(2e^(2x)-2)/((e^(2x)-1)+2xe^(2x))`
[Using L'Hospital's rule]
`lim_(xto0)(4e^(2x))/(2e^(2x)+4xe^(2x))=1`
[Using L'Hospital rule]
Since `f(x)` is continuous at `x=0`
then
`lim_(xto0)f(x)=f(0)implies =f(0)`