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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Limits XII

The value of

lim_{x \to 0} \frac{1}{x ^{3}} \int_{0}^{x} \frac{t l n ( 1 + t )}{t ^{4} + 4} d t

is

(A)0
(B) $\frac{1}{12}$ ✓ Correct
(C) $\frac{1}{24}$
(D) $\frac{1}{64}$

Answer: (B)

Solution

Let

f (t) = \frac{t l n ( 1 + t )}{t ^{4} + 4}

. As

t \to 0

:

ln (1 + t) \sim t - \frac{t ^{2}}{2} + O (t^{3})

f (t) \sim \frac{t \cdot t}{4} = \frac{t ^{2}}{4}

Using L'Hôpital's rule (or Taylor expansion):

x \to 0 lim \frac{1}{x ^{3}} \int_{0}^{x} f (t) d t = x \to 0 lim \frac{f ( x )}{3 x ^{2}}

Now:

x \to 0 lim \frac{x ln ( 1 + x )}{3 x ^{2} ( x ^{4} + 4 )} = x \to 0 lim \frac{ln ( 1 + x )}{3 x ( x ^{4} + 4 )}

Using

ln (1 + x) \sim x

:

= x \to 0 lim \frac{x}{3 x \cdot 4} = \frac{1}{12}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option

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