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.
 A: Let $t=x-\frac{\pi}{2}$, then we have to compute
$$\lim_{t \to 0}\frac{1}{t}-\cot t.$$
Now use the series expansion for $\cot t$ given by
$$\cot t=\frac{\cos t}{\sin t}=\left(1-\frac{t^2}{2}+o(t^3)\right)\frac{1}{t}\left(1+\frac{t^2}{6}+o(t^3)\right)=\frac{1}{t}-\frac{t}{3}+o(t).$$
So
$$\lim_{t \to 0}\frac{1}{t}-\cot t=\lim_{t \to 0}\frac{t}{3}+o(t)=0.$$
A: Sub $u=\frac{\pi}{2}-x$ then we have
$$
\begin{align}
&\lim_{u\to0} \frac{1}{u}-\cot u \\
&=\lim_{u\to0} \frac{1-u\cot u}{u} \\
&=\lim_{u\to0} \left(-\cot u+u\csc^2u \right) \\
&=\lim_{u\to0} \left(\frac{u}{\sin^2u}-\frac{\cos u}{\sin u} \right) \\
&=\lim_{u\to0} \frac{u-\sin u\cos u}{\sin^2u} \\
&=\lim_{u\to0} \frac{u-\sin u\cos u}{2\sin u\cos u} \\
&=\lim_{u\to0} \left(\frac{u}{2\sin u\cos u} - \frac{1}{2} \right) \\
&=0
\end{align}
$$
A: Let $y= \dfrac{\pi}{2}-x$
And rewrite as 
$\lim_y \rightarrow 0 \dfrac{\tan y - y}{y^2} \dfrac{y}{\tan y}$
A: 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.
