We have lim(x→0) (sinx + cosx)cosec x
= lim (x→0–) (1 + (sinx + cosx – 1))cosec x
= elim (x→0–) (sinx + cosx – 1)/sinx
Since f(x) is continuous at x = 0, so
lim (x→0+) f(x) = lim (x→0–) f(x) = f(0)
1/b = e = a
Thus, a = e, b = 1/e
Hence, the value of {e(a + b) + 2}
= {e( e + 1/e ) + 2}
= (e2 + 3)