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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

`log2`

C

`log4`

D

`e^(4)`

Text Solution

Verified by Experts

The correct Answer is:
C
`lim_(xto0)(2^(x)-2^(-x))/x=lim_(xto0)2^(x)log2`
`+2^(-x)log2` [by L'Hospital's rule]
`=log2+log2=log4`
Since, function is continuous at
`x=0`
`:.f(0)=lim_(xto0)(2^(x)-2^(-x))/x`
`impliesf(0)=log4`
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