How do I evaluate this trigonometric limit: $\lim_{x\to\frac\pi4}\frac{\sqrt2\cos(x)-1}{\cot(x)-1}$? $$\lim\limits_{x\to\frac\pi4}\frac{\sqrt2\cos(x)-1}{\cot(x)-1}=?$$
I tried using many formulae but am not able to cancel out the zero factor. Need help...thanks!!
 A: Let $f(x) = \sqrt 2 \cos x, g(x) = \cot x.$ The expression equals
$$\frac{f(x) - f(\pi/4)}{g(x) - g(\pi/4)} = \frac{(f(x) - f(\pi/4))/(x-\pi/4)}{(g(x) - g(\pi/4))/(x-\pi/4)}.$$
By definition of the derivative, as $x\to \pi/4,$ the numerator on the right $\to f'(\pi/4),$ the denominator $\to g'(\pi/4).$ The desired limit is thus $f'(\pi/4)/g'(\pi/4).$ This is a simple computation.
A: Hint:
$$\cot x-\cot45^\circ=?$$
Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html for $$\cos x-\cos45^\circ=?$$
A: Hint: let $y=x-\frac{\pi}{4}$ so that $y \to 0$ when $x\to \frac{\pi}{4}$. Then:
$$
\cos(x)=\cos(y+\frac{\pi}{4})= \frac{1}{\sqrt{2}}(\cos(y) -\sin(y)) \\
\cot(x)=\frac{\cos(y+\frac{\pi}{4})}{\sin(y+\frac{\pi}{4})} = \frac{\cos(y) -\sin(y)}{\cos(y) +\sin(y)}
$$
The limit reduces to (using just the book definitions of derivatives at the last step):
$$
\begin{align}
\lim_{y\to0}\frac{\cos(y)-\sin(y)-1}{\frac{\cos(y)-\sin(y)}{cos(y)+sin(y)}-1} & = \lim_{y\to0}\frac{(\cos(y)-\sin(y)-1)(\cos(y)+\sin(y))}{-2 \sin(y)} \\
& = \frac{-1}{2} \lim_{y\to0} \left( \,\frac{\cos(y) - 1}{y}\,\frac{y}{\sin(y)}- 1\right)(\cos(y)+\sin(y)) \\
& = \frac{-1}{2}\left(\cos'(0)\,\frac{1}{\sin'(0)}-1\right)(\cos(0)+\sin(0)) \\
 & = \;\cdots
\end{align}
$$
A: In the same spirit as @dxiv's answer, let $x=y+\frac \pi 4$; so $$\sqrt 2\cos(x)=\cos (y)-\sin (y)\implies \sqrt 2\cos(x)-1=\cos (y)-\sin (y)-1$$ $$\cot(x)=\frac{\cos(y)-\sin(y)}{\cos(y)+\sin(y)}\implies\cot(x)-1=-\frac{2 \sin (y)}{\sin (y)+\cos (y)}$$ $$\frac{\sqrt2\cos(x)-1}{\cot(x)-1}=\frac{1}{2} \left(1+\sin (y)+\cos (y) \tan \left(\frac{y}{2}\right)\right)$$
A: I think that you know about L'Hospital's rule. 
Let $\ f(x) = \sqrt{2}\cos{x}-1\ $,$\ g(x) = \cot{x} -1\ $
Then derivatives of this functions are:
$\ f^{'}(x) =-\sqrt{2}\sin{x} $, $ \ g(x) = \large - \frac {1}{sin^{2}{x}}\ $
Limit is:
$ lim_{x \to \frac{\pi}{4}}\frac {f^{'}(x)}{g^{'}(x)}=\lim_{x \to \frac{\pi}{4}}\sqrt{2}\sin^{3}{x} = \frac {1}{2}\ $
