GRE (2)

Calculus Level 3

Evaluate the following limit:

lim ⁡ x → 0 ( 1 x 2 ∫ 0 x t + t 2 1 + sin ⁡ t d t ) \lim_{x\to\ 0} \Bigg(\frac {1}{x^2}\int_{0}^{x} \frac{t+t^2}{1+\sin t} dt\Bigg)

1 2 \frac{1}{2} 1 π \frac{1}{\pi} 1 2 π \frac{1}{2\pi} π 2 \frac{\pi}{2} 1 1

Hana Wehbi by Hana Wehbi , 51, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hana Wehbi
May 5, 2018

The integral equals to 0 when x = 0 x=0 , then the limit is of the indeterminate form 0 0 \frac{0}{0} , so we can apply L'Hopital's rule:

lim ⁡ x → 0 ( 1 x 2 ∫ 0 x t + t 2 1 + sin ⁡ t d t ) = lim ⁡ x → 0 x + x 2 1 + sin ⁡ x 2 x = lim ⁡ x → 0 x ( 1 + x ) 2 x ( 1 + sin ⁡ x ) = lim ⁡ x → 0 1 + x 2 ( 1 + sin ⁡ x ) = 1 2 \lim_{x\to\ 0} \Bigg(\frac {1}{x^2}\int_{0}^{x} \frac{t+t^2}{1+\sin t} dt\Bigg)\ = \lim_{x \to\ 0} \frac {\frac{x+x^2}{1+\sin x}}{2x} = \lim_{x\to\ 0} \frac {x(1+x)}{2x(1+\sin x)} = \lim_{x\to \ 0} \frac {1+x}{2(1+\sin x)} =\frac{1}{2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...