Limits at infinity involving trigonometry

Question

I was wondering as I stumbled upon a question like this: $$ \lim_{x\to(\pi/2)^+}\frac{\tan(x)-x}{\tan(2x)+3} $$

$\tan(x)$, when $x$ is approaching $\pi/2$ is $-∞$, and I know that from the graph, but is there any other way that this can be done without looking at the graph? Maybe using a calculator, or drawing a table?

EDIT: I would also like to ask why is $\tan(2x) = 0$ and not $-∞$ like $\tan(x)$.

Answer 1

use that $$\tan(2x)=\frac{2\tan(x))}{1-\tan(x)^2}$$ and $$\tan(x)=\frac{\sin(x)}{\cos(x)}$$

Answer 2

Tan (2x) evaluates the tangent of 2*x, not of x. Thus, tan lim tan(2x) as x-> pi/2 evaluates tan(pi), which = 0. Also, as Dr. Graubner said in his answer, use the trig identities to rewrite the function to find the proof analytically.