I had calculated $\lim_{z\to 1} (1-z).\tan(πz/2)$ by L'Hospital Rule as Follows I took $\tan(πz/2)$ to the denominator of the denominator i.e $(1-z)/1/\tan(πz/2)$ which becomes $(1-z)/\cot(πz/2)$ and after applying L hospital rule the limit value is $2/π$. Same can be handled by substituting $z = 1 - x$
However, when I tried the same by taking $(1-z)$ down instead of $\tan(πz/2)$, the limit became almost unsolvable
My questions are
- does taking a different function to the denominator of denominator make the limit unsolvable
- is there a way to predict/guess which function should remain at top (say $f$) and which function to take down (say $g$, so $f/1/g$) for evaluating the limit easily
- Are there any other ways/methods to solve it?
Q1: does taking a different function to the denominator of denominator make the limit unsolvable?
No. $$\lim_{z \to 1}(1−z)\tan \tfrac12 \pi z =\lim_{z \to 1}\frac{\tan \frac12\pi z}{\frac{1}{1-z}}=\lim_{z \to 1}\frac{\frac12 \pi \sec^2 \frac12 \pi z}{\frac{1}{(1-z)^2}}$$ $$=\lim_{z \to 1}\frac{\tfrac12 \pi (1-z)^2}{ \cos^2 \tfrac12 \pi z}=\tfrac12 \pi\left(\lim_{z \to 1}\frac{ 1-z}{ \cos \frac12 \pi z}\right)^2=\tfrac12 \pi\left(\lim_{z \to 1}\frac{-1 }{ -\frac12 \pi \sin \frac12 \pi z}\right)^2=\frac{2}{\pi}$$
Q2: is there a way to predict/guess which function should remain at top (say f) and which function to take down (say g, so f/1/g) for evaluating the limit easily?
As per the comment, you can look ahead and try to guess. I don't know of anything better than that.
Q3: Are there any other ways/methods to solve it?
As you suggest in the question, you could do this: $$\lim_{z \to 1}(1−z)\tan \tfrac12 \pi z =\lim_{x \to 0}x \cot \tfrac12\pi x=\lim_{x \to 0}x \frac{\cos \tfrac12\pi x}{\sin \tfrac12\pi x}$$ $$=\lim_{x \to 0} \cos \tfrac12\pi x \; \lim_{x \to 0} \frac{ x}{\sin \tfrac12\pi x} = 1 \times\frac{2}{\pi} $$