$\lim_{x\to\ \frac{\pi}{2}^-}\frac{1}{x-\frac{\pi}{2}}+\tan{x} $

Question

$$\lim_{x\to\ \frac{\pi}{2}^-}\frac{1}{x-\frac{\pi}{2}}+\tan x $$

Can anyone give me a hint I'm struggle.

Edit maybe I should say what I've done:

I used the following change of variable: $t=x-\frac{\pi}{2}$ so I got:

$$\lim_{t\to\ 0^-}\frac{1}{t}+\tan \left(t+\frac{\pi}{2}\right)$$

$$\lim_{t\to\ 0^-} \frac{1+t\tan\left(t+\frac{\pi}{2}\right)}{t} $$

I used B.H and I got something like $\frac{1}{0}$ so it's undefined...

Answer 1

As you've already found, the limit is $\lim_{y\to 0^+}(\cot y -\frac{1}{y})$. For small $y$, $\cot y=\frac{\cos y}{\sin y}\approx\frac{1}{y}\frac{1-y^2/2}{1-y^2/6}\approx\frac{1}{y}(1-y^2/3)$ so $\cot y -\frac{1}{y}\approx-\frac{y}{3}$. The limit is therefore $0$.

Answer 2

Thanks for the trigonometric identity: $\tan({x+\frac{\pi}{2})} = -\cot{x}$

Answer 3

Usually when it comes to L'Hospital's Rule, it is easier to take derivatives of sines and cosines, instead of tangents and cotangents, particularly when they are in fractions, like here. So if you write in your given limit $tanx=\frac{sinx}{cosx}$, and adding fractions together, you get $\lim_{x\to\ \frac{\pi}{2}^-}\frac{cosx+(x-\pi/2)sinx}{(x-\pi/2)cosx} $. This is a zero over zero situation for which the Rule can be applied. While taking individual derivatives is easy, be ready to apply the Rule successively.(Patience!). Ultimately you will find the answer. NOTE The substitution as suggested by Lord.....is a better approach.