If x 0 ( xx)^ 1x^2 = p, then 96 e p is equal to — Solution | JEE Mathematics

NVAMediumJEE 2025Limits

If $\lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p$ , then $96 \log_e p$ is equal to _____

Answer

Correct answer:32

Standard Method

Given:

p = \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}}

Find: $96 \log_e p$

The given limit is of the indeterminate form $1^\infty$ , so logarithmic manipulation is used.

Let

y = \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}}

Then,

\log y = \frac{1}{x^2} \log\left( \frac{\tan x}{x} \right)

Now evaluate the limit using the Taylor expansion

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

So,

\frac{\tan x}{x} = 1 + \frac{x^2}{3} + \cdots

Hence,

\log\left( \frac{\tan x}{x} \right) = \log\left(1 + \frac{x^2}{3}\right) \approx \frac{x^2}{3}

Therefore,

\lim_{x \to 0} \frac{\log\left( \frac{\tan x}{x} \right)}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{3}}{x^2} = \frac{1}{3}

Thus,

\log y = \frac{1}{3}

and so,

y = e^{\frac{1}{3}}

Therefore,

p = e^{\frac{1}{3}}

Now,

96 \log_e p = 96 \log_e\left(e^{\frac{1}{3}}\right) = 96 \cdot \frac{1}{3} = 32

Therefore, the required value is $32$ .

Using the $$1^\infty$$ transformation

Given:

p = \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}}

Find: $96 \log_e p$

For a limit of the form

\lim [f(x)]^{g(x)} = e^{\lim g(x)(f(x)-1)}

when $f(x) \to 1$ , we get

\log_e p = \lim_{x \to 0} \frac{1}{x^2} \left( \frac{\tan x}{x} - 1 \right)

This becomes

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

Using

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

we have

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

So,

\log_e p = \lim_{x \to 0} \left( \frac{1}{3} + \frac{2x^2}{15} + \dots \right) = \frac{1}{3}

Hence,

96 \log_e p = 96 \times \frac{1}{3} = 32

Therefore, the required value is $32$ .

Treating the expression directly as $1$ because $\frac{\tan x}{x} \to 1$ . This is wrong because the exponent $\frac{1}{x^2} \to \infty$ also matters, giving the indeterminate form $1^\infty$ . Use logarithms or the standard transformation for exponential limits.
Using an incomplete expansion such as $\tan x \approx x$ only. This removes the first non-zero correction term and makes the limit vanish incorrectly. Retain the next term: $\tan x = x + \frac{x^3}{3} + \cdots$ .
Replacing $\log(1+u)$ by $u$ without checking what $u$ is. The approximation is valid here only because $u = \frac{x^2}{3} \to 0$ . First identify the small quantity, then apply $\log(1+u) \approx u$ .

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