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The value of f at x =0 so that funcation...

The value of f at x =0 so that funcation ` f(x) = (2^(x) -2^(-x))/x , x ne 0` is continuous at x =0 is

A

0

B

log 2

C

4

D

log 4

Text Solution

Verified by Experts

The correct Answer is:
D
`lim_(x to 0)(2^(x)-2^(-x))/(x)=lim_(x to0)((2^(x)log2+2^(-x)log2))/(1) " " `[by L'Hospital's rule]
`=log2+log2=log4`
Since, the function is continuous at x = 0
`therefore f(0)=lim_(x to 0)(2^(x)-2^(-x))/(x)`
`rArr f(0) log 4`
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