LIMITS : $\lim_{x\to\ -1} \frac{\cos(2)-\cos(2x)}{(x)^2-|x|}$ $$\lim_{x\to\ -1} \frac{\cos(2)-\cos(2x)}{(x)^2-|x|}$$
I am not able to get the answer , Here is my solution :
i have replaced $x$ by $h-1$
$$\lim_{h\to0} \frac{\cos(2)-\cos(2(h-1))}{((h-1)^2)+(h-1)}$$
so i get : $$\lim_{h\to0} \frac{1}{2}2.\sin(h-2)\sin h$$
i have used $$\lim_{h\to0} \frac{\sin(h)}{h} = 1$$
 A: Since $\cos 2x$ is differentiable, $\lim_{x\to -1}\frac{\cos 2x -\cos 2}{x+1}$ exists and its value is $2\sin2$. Thus
\begin{align}
\lim_{x\to -1}\frac{\cos 2-\cos 2x}{x^2-|x|}&=\lim_{x\to-1}\left(\frac{\cos 2x - \cos 2}{x+1}\cdot \frac{-1}{x}\right)\\
&=\lim_{x\to-1}\frac{\cos 2x - \cos 2}{x+1}\lim_{x\to-1}\frac{-1}{x}\\
&=2\sin2\cdot 1\\
&=2\sin 2.
\end{align}
A: Hint: Since you are looking for the limit around $x=-1$ (hence, in the negative axis), you simply assume $|x|=x$ and use L'Hopital:
$$\lim_{x\to\ -1} \frac{\cos(2)-\cos(2x)}{(x)^2-|x|} = \lim_{x\to\ -1} \frac{\cos(2)-\cos(2x)}{x^2+x} = ... $$
A: Your method is sound: you can assume to work in a neighborhood of $-1$ like $(-2,0)$, so $|x|=-x$. Then your substitution gives
\begin{align}
\lim_{h\to0}\frac{\cos(2)-\cos(2h-2)}{(h-1)^2+h-1}
&=\lim_{h\to0}\frac{\cos(2)-\cos(2h)\cos(2)-\sin(2h)\sin(2)}{h(h-1)}
\\[6px]
&=\lim_{h\to0}\frac{2\cos(2)}{h-1}\frac{1-\cos(2h)}{2h}-
\lim_{h\to0}\frac{2\sin(2)}{h-1}\frac{\sin(2h)}{2h}
\end{align}
The limit of the second factor in the first summand is known to be $0$; the second summand is easily seen to have limit $2\sin(2)$
A: Let $f(x) = \cos 2x, g(x) = x^2 + x.$ Then for $x$ close to $-1$ the expression equals
$$-\frac{f(x) - f(-1)}{g(x) - g(-1)} = -\frac{(f(x) - f(-1))/(x-(-1))}{(g(x) - g(-1))/(x-(-1))}.$$
By definition of the derivative, this goes to $-f'(-1)/g(-1),$ which is simple to compute.
