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JEE Main · 2016 · 03 Apr 2016 · Offline · Limits

Let

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

then

lo g p

is equal to:

(A) $1$
(B) $\frac{1}{2}$ ✓ Correct
(C) $\frac{1}{4}$
(D) $2$

Answer: (B)

Solution

Take logarithms of

p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{1/ (2 x)}

:

lo g p = lim_{x \to 0^{+}} \frac{1}{2 x} lo g (1 + tan^{2} x)

. For small arguments

lo g (1 + u) \approx u

, so this is

lim_{x \to 0^{+}} \frac{t a n ^{2} x}{2 x}

, and since

tan x \approx x

, it becomes

lim_{x \to 0^{+}} \frac{x}{2 x} = \frac{1}{2}

. Hence

lo g p = \frac{1}{2}

.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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