UPSC NDA Math Model Paper 5

If we substitute x = 0 then we observe that the given limit is of the

0

0

form.
Thus we will differentiate numerator and denominator and write the given limit as:

lim

x→0

(1+x)n−1x=

lim

x→0

n(1+x)n−1−0

1

=n

lim

x→0

(1+x)n−1
=n(1+0)n−1
=n
Therefore,

lim

x→0

(1+x)n−1

x

=n