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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

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Verified by Experts

The correct Answer is:
D
`lim_(xto0) (2^(x)-2^(-x))/x = lim_(x to 0) 2^(x) log 2 + 2^(-x) log 2` [ by L' Hospital's rule ]
= log 4 ltbr Since, the function is continuous at x=0 .
` f(0)= lim_(xto 0) f(x)`
f(x) = log 4
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