Sine over sine

Calculus Level 4

lim x π 2 sin x ( sin x ) sin x 1 sin x + ln ( sin x ) \large \lim_{x\rightarrow \frac{\pi}{2}} \dfrac{ \sin x - (\sin x)^{\sin x} }{1-\sin x + \ln (\sin x)}

Find the value of the limit above.


The answer is 2.

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

Chew-Seong Cheong
Apr 10, 2017

Relevant wiki: L'Hopital's Rule - Basic

L = lim x π 2 sin x sin sin x x 1 sin x + ln ( sin x ) Let u = sin x = lim u 1 u u u 1 u + ln u A 0/0 cases, L’H o ˆ pital’s rule applies. = lim u 1 1 ( ln u + 1 ) u u 1 + 1 u Again a 0/0 case = lim u 1 u u 1 ( ln u + 1 ) 2 u u 1 u 2 Differentiating up and down w.r.t. u again = 2 \begin{aligned} L & = \lim_{x \to \frac \pi 2} \frac {{\color{#3D99F6}\sin x} - {\color{#3D99F6}\sin^{\sin x} x}}{1-{\color{#3D99F6}\sin x} + \ln ({\color{#3D99F6}\sin x} )} & \small \color{#3D99F6} \text{Let }u = \sin x \\ & = \lim_{u \to 1} \frac {{\color{#3D99F6}u}-\color{#3D99F6}u^u}{1-{\color{#3D99F6}u}+\ln {\color{#3D99F6}u}} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{u \to 1} \frac {1-(\ln u+1)u^u}{-1+\frac 1u} & \small \color{#3D99F6} \text{Again a 0/0 case} \\ & = \lim_{u \to 1} \frac {-u^{u-1} - (\ln u+1)^2u^u}{-\frac 1{u^2}} & \small \color{#3D99F6} \text{Differentiating up and down w.r.t. }u \text{ again} \\ & = \boxed{2} \end{aligned}

Yes, did the same way, but i didnt converted sin x = u \sin x = u .

I struggled with sin x \sin x

Md Zuhair - 4 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...