Trigonometric Limit 0*infinity

Question

I came across this trigonometric limit. I’ve tried to solve it by transformation but failed.

How can I evaluate it? Answer is $2/\pi$.

$$ \lim_{x\to 1} (1-x) \tan\left(\frac{x\pi}{2}\right) $$

Answer 1

You want to use the fact that $\lim_{\theta\to 0} \frac{\sin\theta}{\theta} = 1$. \begin{align*} \lim_{x\to 1}(1-x) \tan\left(\frac{x\pi}{2}\right) &= \lim_{x\to 1}(1-x)\frac{\sin\left(\frac{x\pi}{2}\right)}{\cos\left(\frac{x\pi}{2}\right)} = \lim_{x\to 1}\frac{1-x}{\cos\left(\frac{x\pi}{2}\right)} \end{align*} since the sine term tends to one. Let $t = 1-x$. Then $t\to 0$ as $x\to 1$ and \begin{align*} \lim_{x\to 1}\frac{1-x}{\cos\left(\frac{x\pi}{2}\right)} &= \lim_{t\to 0} \frac{t}{\cos\left(\frac{\pi}{2} - \frac{t\pi}{2}\right)} \\&= \lim_{t\to 0} \frac{t}{\sin(t\pi/2)} \\&= \lim_{t\to 0} \frac{2}{\pi} \cdot \frac{t\pi/2}{\sin(t\pi/2)} \end{align*} Now let $\theta = \frac{t\pi}{2}$ and you are home free.

Alternatively, picking up on the good Doctor's suggestion, let $f(x) = \cot\left(\frac{\pi x}{2}\right)$. Then $$ f'(x) = -\csc^2\left(\frac{\pi x}{2}\right)\cdot \frac{\pi}{2} $$ So $f'(1) = - \frac{\pi}{2}$. Your limit can be expressed as $$ \lim_{x\to 1} \left[(-1)\cdot \frac{x-1}{f(x)-f(1)}\right] = (-1) \cdot \frac{1}{f'(1)} = \frac{2}{\pi} $$

Answer 2

write it in the form $$\frac{1-x}{\cot\left(\frac{x \pi}{x}\right)}$$ and use L'Hospital the result should be $$\frac{2}{\pi}$$

Answer 3

Let $x=1-y.$ Then $$(1-x)\tan (\pi x/2)=y\tan (\pi /2-y\pi /2)=y\cot (y\pi /2)=$$ $$=\frac {2}{\pi}\cdot \frac {y\pi /2}{\sin (y\pi /2)}\cdot \cos (y\pi /2).$$ Now $y\pi /2\to 0$ as $x\to 1 .$ So $\frac {y\pi /2}{\sin (y\pi /2)}\to 1$ and $\cos (y\pi /2)\to 1$ as $x\to 0.$

Answer 4

When $x\to 1$, $x\pi/2\to\pi/2$. Set $x\pi/2=\pi/2-t$, so $$ x=1-\frac{2t}{\pi} $$ and the limit becomes $$ \lim_{t\to0}\frac{2t}{\pi}\tan\left(\frac{\pi}{2}-t\right) = \frac{2}{\pi}\lim_{t\to0}t\cot t= \frac{2}{\pi}\lim_{t\to0}\frac{t}{\sin t}\cos t $$