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The function f:""R""~""{0}vecR given ...

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

A

2

B

`-1`

C

0

D

1

Text Solution

Verified by Experts

The correct Answer is:
D
`lim_(x to 0){(1)/(x)-(2)/(e^(2x)-1)}=lim_(x to 0)(e^(2x)-1-2x)/(x(e^(2x)-1))" "[(0)/(0)"form"]`
` " " `[by L' Hospital's rule]
`=lim_(x to 0)(2e^(2x)-2)/((e^(2x)-1)+2xe^(2x)) " " `[again by using L' Hospitals rule]
`=lim_(x to 0)(4e^(2x))/(4e^(2x)+4xe^(2x))=1`
`because `f(x) is continuous at x = 0, then
`lim_(x to 0) f(x)=f(0)`
`therefore 1 =f(0)`
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