Answer 1

Since f is continuous at x = 0.
`lim_( x -> 0 ) f(x) = f(0) `
Given f(0) = k
∴ `lim_( x -> 0) f(x) = k`                    (i)
Now `lim_( x -> 0) f(x) = lim_( x -> 0 ) [(e^x  - 1)sinx]/[x^2]`
                                   = `lim_( x -> 0) ([e^x - 1]/[x])([sin x]/[x])`
                                   = `lim_( x -> 0) ([e^x - 1]/[x]) lim_(x->0)([sin x]/[x])`
                                   = log e x 1
                                   = 1 x 1
`therefore lim_( x -> 0 ) = f(x) = 1`               (ii)
from (i) and (ii)
k = 1