Test Index

UPSC NDA I 2018 Mathematics Solved Paper

Question : 28 of 120

Marks: +1, -0

What is

\underset{x→0}{lim}\frac{tanx}{sin2x}

equal to?

Solution:

Concept:

The limit

\underset{x→0}{\lim}\frac{tan x}{\sin 2x}

is evaluated using the standard limit

\underset{x→0}{\lim}\frac{tan x}{x}=1

and

\underset{x→0}{\lim}\frac{\sin kx}{kx}=1

Explanation:

Rewrite the expression by factoring

x

and

2x

in numerator and denominator.

\underset{x→0}{\lim}\frac{tan x}{\sin 2x} = \underset{x→0}{\lim} \frac{\frac{tan x}{x} \cdot x}{\frac{\sin 2x}{2x} \cdot 2x} = \underset{x→0}{\lim} \frac{\frac{tan x}{x}}{\frac{\sin 2x}{2x}} \cdot \frac{x}{2x} = \frac12 \cdot \underset{x→0}{\lim} \frac{\frac{tan x}{x}}{\frac{\sin 2x}{2x}}

Since the limit of a quotient equals the quotient of the limits (when denominator limit is non-zero), we have:

\frac12 \cdot \frac{\underset{x→0}{\lim}\frac{tan x}{x}}{\underset{x→0}{\lim}\frac{\sin 2x}{2x}} = \frac12 \cdot \frac{1}{1} = \frac12

Thus, the required limit is

\frac12

Answer:

\frac12

(Option A)

Go to Question:

Prev Question

Next Question