Test Index

PUNE MCA Exam 2014 Question Paper with answers for online practice

© examsnet.com

Question : 42 of 58

Marks: +1, -0

\lim\limits_{x\to 0}(1+3x)^{1/x}

$e^{-3}$
$e^{3}$
3
-3

Solution:

Concept:

Use the standard limit

\lim_{x\to 0}(1+ax)^{1/x}=e^a

.

Explanation:

The given limit is

\lim_{x\to 0}(1+3x)^{1/x}

.

This is of the form

1^\infty

.

Using the known result

\lim_{x\to 0}(1+ax)^{1/x}=e^a

, with

a=3

, we get the limit as

e^3

.

Alternatively, rewrite as

\lim_{x\to 0} e^{\frac{1}{x}\ln(1+3x)} = e^{\lim_{x\to 0}\frac{\ln(1+3x)}{x}}

.

Since

\ln(1+3x)\sim 3x

as

x\to0

, the limit inside is

3

, giving

e^3

.

Answer:

e^3

© examsnet.com

Go to Question:

Prev Question

Next Question