Test Index

UPSC NDA Math Model Paper 5

© examsnet.com

Question : 27 of 120

Marks: +1, -0

What is

\lim\limits_{x \to 0} \frac{(1+x)^{n}-1}{x}

equal to?

0
1
$n$
$n - 1$

Solution:

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

\frac{0}{0}

form.

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

\lim\limits_{x \to 0} ((1+x)^{n} - 1) x = \lim\limits_{x \to 0} \frac{n(1+x)^{n-1}-0}{1}

= n \lim\limits_{x \to 0} (1+x)^{n-1}

= n (1+0)^{n-1}

= n

Therefore,

\lim\limits_{x \to 0} \frac{(1+x)^{n}-1}{x} = n

© examsnet.com

Go to Question:

Prev Question

Next Question