# Thread: Given function, evaluate limit?

1. ## Given function, evaluate limit?

If $f(x) = \lim_{n\rightarrow \infty} \left\{\sin x + 2\sin^2 x + 3\sin^3 x + \mbox{...} + n\sin^n x\right\}$, then evaluate:

$\lim_{x\rightarrow \frac{\pi}{2}} \left\{(1 - \sin x)^2 f(x)\right\}^{\frac{1}{(sin x - 1)}}$.

2. ## Telescoping

Well, first notice that $(1-\sin x)^2f(x)$ is a telescoping series converging on simply $\sin x$. So to evaluate $L=\lim_{x\rightarrow \frac\pi2} (\sin x)^{\frac1{\sin x-1}}$ , we first substitute $y=\sin x$ , making $L=\lim_{y\rightarrow0}y^{\frac1{y-1}}$ and then $z=\frac1{y-1}$, so $L=\lim_{z\rightarrow\infty}(1+\frac1z)^z=e$

3. $1+\sin x+\sin^2 x+\sin^3 x+.........=\frac{1}{1-\sin x}$

Differentiating w.r.t x

$\cos x+2\sin x\cos x+3\sin^2 x\cos x+4\sin^3 x\cos x+.....=\frac{\cos x}{(1-\sin x)^2}$

$1+2\sin x+3\sin^2 x+4\sin^3 x+.....=\frac{1}{(1-\sin x)^2}$

Multiplying throughout by sinx

$
\sin x+2\sin^2 x+3\sin^3 x+4\sin^4 x+.....=\frac{\sin x}{(1-\sin x)^2}
$

$
(1-\sin x)^2f(x)=\sin x
$

$\lim_{x\to \frac{\pi}{2}}((1-\sin x)^2f(x))^{\frac{1}{\sin x-1}}
$

$=e^{\lim_{x\to \frac{\pi}{2}}((1-\sin x)^2f(x)-1){\frac{1}{\sin x-1}}}$

=e