Evaluating $\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x}$

Question

My sister is a $11$'th grade student and she is preparing for a Calculus test paper. This was in last year's sample exam : $$\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x}$$

She asked for my help but really gave me some really crippling conditions. The solution is quite easy to reach if one uses L'Hospital, Taylor series, or a really plain $y - \pi= x$ substitution.

But they weren't even taught substitutions yet ! So, how would you go about this ? Maybe some clever trigonometric massaging?

A basic manipulations yields:

$$\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x} = \lim\limits_{x \to \pi} (\pi - x ) \tan\bigg(\frac{x}{2}\bigg)$$

Any ideas?

Answer 1

Hint:

1) Multiply numerator and denominator of the fraction by $(1-\cos x)$ and note that $1-\cos^2 x=\sin^2 x $.

2) use $ \sin x=\sin(\pi-x)$

3) use the special limit $$\lim_{x \to 0}\frac{\sin x}{x}=1$$

Answer 2

\begin{align} &\lim\limits_{x \to \pi} \frac{(\pi -x)\sin x}{1 + \cos x}\\ =&\lim\limits_{x \to \pi} \frac{-\sin x}{\frac{\cos x-(-1)}{x-\pi}}\\ =&\lim\limits_{x \to \pi} \frac{-\sin x}{\frac{\cos x-\cos \pi}{x-\pi}}\\ =&\lim\limits_{x \to \pi} \frac{-\sin x}{\cos' \pi}\\ =&\lim\limits_{x \to \pi} \frac{-\sin x}{-\sin \pi}\\ =&\frac{\sin \pi}{\sin \pi}\\ =&1\\ \end{align}

Answer 3

Use

$$\frac{(\pi -x)\sin x}{1 + \cos x}=(\pi-x)\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{\cos^2\left(\frac x2\right)}$$

Answer 4

Check the solution and do tell in case of any doubt.

Answer 5

Set $\theta =\pi -x \iff x=\pi-\theta$. Then $$\frac{(\pi-x)\sin x}{1+\cos x}=\frac{\theta\sin \theta}{1-\cos \theta}\sim_0\frac{\theta^2}{\dfrac{\theta^2}2}=2.$$ Thus the limit is $\color{red}{2}$.

Answer 6

$\lim\limits_{x\rightarrow\pi}\dfrac{(\pi-x)\sin x}{1+\cos x}=\lim\limits_{y\rightarrow 0}\dfrac{y\sin(\pi-y)}{1+\cos(\pi-y)}=\lim\limits_{y\rightarrow 0}\dfrac{y\sin y}{1-\cos y}=\lim\limits_{y\rightarrow 0}\dfrac{2y\sin {\dfrac y2}\cos {\dfrac y2}}{2\sin^2 {\dfrac y2}}=\lim\limits_{y\rightarrow 0}\left(\dfrac {\sin {\dfrac y2}}y\right)^{-1}\dot\cos{\dfrac y2}=1.$