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If f(x)={sinx ,x!=npi,n in I2,ot h e r ...

If `f(x)={sinx ,x!=npi,n in I2,ot h e r w i s e` `g(x)={x^2+1,x!=0,4,x=0 5,x=2` then `("lim")_(xvec0)g{f(x)}i s=`

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The correct Answer is:
1
`underset(xto0^(+))limg{f(x)}=g(f(0^(+)))=g((sin0^(+)))=g(0^(+))=(0)^(2)+1=1`
`underset(xto0^(-))limg{f(x)}=g(f(0^(-)))=g((sin0^(-)))=g(0^(-))=(0)^(2)+1=1`
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