UPSC NDA Math Model Paper 10

lim

x→2

sin(ex−2−1)

log(x−1)

On substituting h=x−2, we get:
=

lim

h→0

sin(eh−1)

log(1+h)

This can be rearranged as:
=

lim

h→0

sin(eh−1)

eh−1

.

eh−1

h

.

h

log(1+h)

=1.1.1
=1 and f(2)=k
∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e. k = 1