NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

The value of $\displaystyle \lim_{x \to 0} \frac{\tan x - x}{x^2 \tan x}$ is equal to:

Choose the correct answer:

Explanation

Solution

We need to evaluate the limit:

L = \lim_{x \to 0} \frac{\tan x - x}{x^2 \tan x}

Step 1: Simplify using standard limits

We know that as $x \to 0$ , $\frac{\tan x}{x} = 1$ . To make use of this, we can multiply and divide the denominator by $x$ :

L = \lim_{x \to 0} \frac{\tan x - x}{x^2 \cdot x \cdot \left(\frac{\tan x}{x}\right)}

Since $\displaystyle \lim_{x \to 0} \frac{\tan x}{x} = 1$ , the expression simplifies to:

L = \lim_{x \to 0} \frac{\tan x - x}{x^3}

Step 2: Check for Indeterminate Form

Substituting $x = 0$ into $\frac{\tan x - x}{x^3}$ gives $\frac{0 - 0}{0} = \frac{0}{0}$ . This is an indeterminate form.

Step 3: Solve using Expansion (Maclaurin Series)

The series expansion for $\tan x$ near $x = 0$ is:

\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots

Substitute this into the limit:

L = \lim_{x \to 0} \frac{\left(x + \frac{x^3}{3} + \dots\right) - x}{x^3}

L = \lim_{x \to 0} \frac{\frac{x^3}{3} + \text{higher order terms}}{x^3}

Step 4: Evaluate the Limit

Divide every term by $x^3$ :

L = \lim_{x \to 0} \left( \frac{1}{3} + \text{terms containing } x \right)

L = \frac{1}{3}

Choose the correct answer:

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