How to evaluate this trigonometric limit? I have been struggling to solve this limit.
What is the limit as $x$ approaches $45^0$ of $$\frac{\sqrt{2}\cos x-1}{\cot x-1}?$$ 
I know how to use L'Hospital's rule to calculate this limit and got the answer as $0.5$. But, how do I calculate the limit by manipulating the function? Please provide only some hints to proceed.
 A: multiply numerator and denominator by $\sqrt{2}\cos x+1$.
you get
$$\frac{2\cos^2 x-1}{(\cot x -1)( \sqrt{2}\cos x+1 )}$$
now $2\cos^2 x-1=\cos 2x$ and $\cos 2x=\frac{1-\tan^2x}{1+\tan^2 x}$
also $\cot x-1=\frac{1-\tan x}{\tan x}$
using all these you get
$$\frac{\sqrt{2}\cos x-1}{\cot x-1}=\frac{\tan x(1+\tan x)}{(1+\tan^2 x)( \sqrt{2}\cos x+1 )}$$
A: Alternative method:
Your expression can be written as:
$$ \lim_{x\to\pi/4} \dfrac{ \sqrt2 \cos x - 1} {\cos x - \sin x} \cdot\sin x $$
$$ = \lim_{h\to 0} \dfrac{ \sqrt2 \cos (h + \pi/4) - 1} {\cos (h + \pi/4) - \sin (h + \pi/4)} \cdot\sin (h + \pi/4) $$
(taking $ x = \pi/4 + h $. And hey, it just looks complicated here! It isn't. )
$$ = \lim_{h\to 0} \dfrac{ (\cos h - \sin h) - 1} {-2\sin h} \cdot\sin (h + \pi/4) $$
(simplifying using the $\sin(A+B), \cos (A+B)$ formulas)
$$ = \lim_{h\to 0} \dfrac{ 1 - \cos h + \sin h} {2\sin h} \cdot\sin (h + \pi/4) $$
$$ = \lim_{h\to 0} \;\;2\sin\left(\frac h 2 \right)\dfrac{ \sin(h/2) + \cos(h/2)} {2 \cdot 2\sin (h/2)\cos(h/2)} \cdot\sin (h + \pi/4) $$
Now, everything breaks down to $0.5$ nicely.
A: HINT 
You have an expression which is of the type 0 / 0. Then, use L'Hospital and you will get your limit.
A: You have
$$
\lim_{x\to\pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}
$$
that you can transform using the substitution $t=x-\pi/4$, that is, $x=t+\pi/4$:
\begin{align}
\lim_{x\to\pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}&=
\lim_{t\to 0}\frac{\sqrt{2}(\frac{1}{\sqrt{2}}\cos t-\frac{1}{\sqrt{2}}\sin t)-1}
  {\frac{\cot t-1}{\cot t+1}-1}\\[2ex]
&=\lim_{t\to 0}(\cos t-\sin t-1)\frac{\cot t+1}{-2}\\[2ex]
&=-\frac{1}{2}\lim_{t\to 0}\frac{(\cos t-\sin t-1)(\cos t+\sin t)}{\sin t}\\[2ex]
\end{align}
You can surely go on from here.
A: $$\lim_{x\to\frac\pi4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\sqrt2\lim_{x\to\frac\pi4}\frac{\cos x-\cos\frac\pi4}{\cot x-\cot\frac\pi4}$$
$$=\sqrt2\frac{\frac{d(\cos x)}{dx}_{(\text{ at }x=\frac\pi4)}}{\frac{d(\cot x)}{dx}_{(\text{ at }x=\frac\pi4)}}=\cdots$$

Alternatively, 
$$F=\lim_{x\to\frac\pi4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\sqrt2\lim_{x\to\frac\pi4}\frac{\cos x-\cos\frac\pi4}{\cot x-\cot\frac\pi4}$$
$$=\sqrt2\lim_{x\to\frac\pi4}\frac{-2\sin\frac{x+\frac\pi4}2\sin\frac{x-\frac\pi4}2}{-\sin(x-\frac\pi4)}\cdot\sin x\sin\frac\pi4$$
$$=\sqrt2\lim_{x\to\frac\pi4}\frac{-2\sin\frac{x+\frac\pi4}2\sin\frac{x-\frac\pi4}2}{-2\sin\frac{x-\frac\pi4}2\cos\frac{x-\frac\pi4}2}\cdot\sin x\sin\frac\pi4$$
As $x\to\frac\pi4, x\ne\frac\pi4\implies \sin\frac{x-\frac\pi4}2\ne0 $
$$\implies F=\sqrt2\lim_{x\to\frac\pi4}\frac{\sin\frac{x+\frac\pi4}2}{\cos\frac{x-\frac\pi4}2}\cdot\sin x\sin\frac\pi4$$
$$=\sqrt2\frac{\sin\frac\pi4}{\cos0}\cdot\sin\frac\pi4\sin\frac\pi4=\cdots$$
