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MHTCET 11 May 2024 Shift 2 Paper

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Question : 111 of 150

Marks: +1, -0

\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}

is

2
-2
$\frac{1}{2}$
$-\frac{1}{2}$

Solution:

Concept:

Use Taylor series expansions of

\tan x

and

\sin x

as

x \to 0

.

Explanation:

The given limit is:

L = \lim_{x \to 0} \frac{x(\tan 2x - 2\tan x)}{(1 - \cos 2x)^2}.

Since

1 - \cos 2x = 2\sin^2 x

, the denominator becomes:

(1 - \cos 2x)^2 = (2\sin^2 x)^2 = 4\sin^4 x.

For small

x

, use the expansions:

\tan 2x = 2x + \frac{(2x)^3}{3} + \cdots = 2x + \frac{8x^3}{3} + \cdots,

2\tan x = 2x + \frac{2x^3}{3} + \cdots.

Subtracting gives:

\tan 2x - 2\tan x = \left(2x + \frac{8x^3}{3} + \cdots\right) - \left(2x + \frac{2x^3}{3} + \cdots\right) = 2x^3 + \cdots.

Also,

\sin x = x + \cdots

, so:

4\sin^4 x = 4x^4 + \cdots.

Therefore:

L = \lim_{x \to 0} \frac{x(2x^3 + \cdots)}{4x^4 + \cdots} = \frac{2x^4}{4x^4} = \frac{1}{2}.

This matches option C.

Answer:

\frac{1}{2}

, i.e. option C.

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