Show that $\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0$. 
Prove that
  $$\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0.$$

I'm not really sure how to proceed. I know that I should not try L'Hôpital's rule (tried that) but not sure how I would incorporate into the Squeeze Theorem or how I would use continuity.
Thanks!
Edit: Turns out I was really dumb and you do use L'Hôpital's rule twice. I made the mistake of differentiating the whole quotient rather than the function on top and the bottom of the vinculum separately.

First note that
$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$
$$\lim_{x\to0}\frac{x}{\sin(x)}=1$$
$$\lim_{x\to0}\frac{1-\cos(x)}{x}=0$$
Keeping these fundamental trigonometric limits in mind, we have
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\left(x-\frac{\pi}{2}\right)}+\tan(x)$$
$$=\lim_{t\to0} \frac{1}{t}-\cot(t)$$
$$=\lim_{t\to0}\frac{\sin(t)-t\cos(t)}{t\sin(t)}$$
$$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)\left(\lim_{t\to0}\frac{\sin(t)-t\cos(t)}{\sin^2(t)}\right)$$
$$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\left(\lim_{t\to0}\frac{t-t\cos(t)}{\sin^2(t)}\right)$$
$$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\left(\lim_{t\to0}\frac{t}{\sin(t)}\right)^2\left(\lim_{t\to0}\frac{1-\cos(t)}{t}\right)$$
Let me know if this helps you.
