The value of limx→0 1x3∫01 tln⁡(1+t)t4+4dt, is

limx→0 1x3∫0x tln⁡(1+t)t4+4dt=limx→0 ∫0x tln⁡(1+t)t4+4dtx3=limx→0 xln⁡(1+x)x4+43x2 [Applying De ‘L’ Hospital's rule]=13limx→0 ln⁡(1+x)xx4+4=13×limx→0 log⁡(1+x)x×1x4+4=13×1×14=112

The value of limx→0 1x3∫01 tln⁡(1+t)t4+4dt, is

limx→0 1x3∫0x tln⁡(1+t)t4+4dt=limx→0 ∫0x tln⁡(1+t)t4+4dtx3=limx→0 xln⁡(1+x)x4+43x2 [Applying De ‘L’ Hospital's rule]=13limx→0 ln⁡(1+x)xx4+4=13×limx→0 log⁡(1+x)x×1x4+4=13×1×14=112

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The value of $lim_{x \to 0} \frac{1}{x^{3}} \int_{0}^{1} \frac{t \ln (1 + t)}{t^{4} + 4} d t,$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$1 / 12$

$1 / 24$

$1 / 64$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

$= lim_{x \to 0} \frac{\int_{0}^{x} \frac{t \ln (1 + t)}{t^{4} + 4} d t}{x^{3}}$

$= lim_{x \to 0} \frac{\frac{x \ln (1 + x)}{x^{4} + 4}}{3 x^{2}}$ [Applying De ‘L’ Hospital's rule]

$\begin{array}{l} = \frac{1}{3} lim_{x \to 0} \frac{\ln (1 + x)}{x (x^{4} + 4)} \\ = \frac{1}{3} \times lim_{x \to 0} \frac{\log (1 + x)}{x} \times \frac{1}{x^{4} + 4} = \frac{1}{3} \times 1 \times \frac{1}{4} = \frac{1}{12} \end{array}$