$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$ I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}}  \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$
the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.
The final result should be $2 \pi + 10$

Let $~f(x) = \left[\tan(\pi x^2)+(2x-1)4^x+6x-4 \right]~$ and 
let $~g(x) = \tan[x - (1/2)].$
Note that $g(x)$ and each term of $f(x)$ are continuous functions.
You want $\lim_{x\to (1/2)} \frac{f(x)}{g(x)}.$
Before applying L'Hopital's rule, first check $f(1/2)$ and $g(1/2)$.
$$f(1/2) = \tan(\pi/4) + (0)4^{(1/2)} + 3 - 4 = 1 + 3 - 4 = 0.$$
$$g(1/2) = \tan(0) = 0.$$
Applying L'Hopital's rule:
$$f'(x)|_{x=(1/2)}
~=~\frac{(2\pi)x}{\cos^2(\pi x^2)} + (2)(4^x) + (2x - 1)\log(4)[4^x] + 6 ~|_{x = (1/2)}
$$
$$= \frac{\pi}{(1/\sqrt{2})^2} + (2)(2) + (0)\log(4)(2) + 6 
~=~ 2\pi + 4 + 0 + 6 ~=~ 2\pi + 10.$$
and
$$g'(x) ~|_{x = (1/2)}
~=~ \frac{1}{\cos^2(x - 1/2)} ~|_{x = (1/2)} = 1.$$
Thus, $$\frac{f'(1/2)}{g'(1/2)} = \frac{2\pi + 10}{1}.$$
