How can find the limit of $(a-x)\tan\frac{\pi x}{2a}$ as $x \to a$?

Question

I tried for an hour but I still couldn't do it. How can I solve this without using L'Hospital's rule?

$$\lim_{x\to a}(a-x)\tan\frac{\pi x}{2a}$$

Answer 1

Assume $a\neq0$.

As $x \to a$, by making the change of variable,
$$ u=\frac{\pi}{2a}(a-x)=\frac{\pi}{2}-\frac{\pi x}{2a} $$ one gets $$ (a-x)\tan\frac{\pi x}{2a}=\frac{2a}{\pi}\frac{u}{\tan u}\to \frac{2a}{\pi}, $$ as $u \to 0$, since in this case $$ \frac{u}{\tan u}=\frac{u}{\sin u} \times \cos u \to 1. $$

Answer 2

you can write $(a-x)\tan\bigg(\dfrac{\pi x}{2a}\bigg)$ as $\dfrac{(a-x)}{\cot\bigg(\frac{\pi x}{2a}\bigg)}$ and limit of this is of form $\frac 00$

hence use l hopital rule

and undermined for is removed and you will get $$\lim_{x\to a}\dfrac{a-x}{\cot(\frac{\pi x}{2a})}=\lim_{x\to a}\dfrac{2a}{\csc^2(\dfrac{\pi x}{2a}).\pi}=\dfrac{2a}{\pi}$$

and hence your answer is $\dfrac {2a}\pi$