calculate the limit + explanation Please help calculation (and understanding) this limit:
$$ \lim_{x\to 1} \frac{x \cos (x-1)-1}{x-1} $$
Not sure if this is the way but I used t = x - 1:
$$ \lim_{t\to 0} \frac{(t+1)\cos(t)-1}{t} $$
and then:
$$ \lim_{t\to 0} \frac{(t+1)\cos(t)-\sin^2(t)-\cos^2(t)}{t} $$
and more or less here I got stuck..Not even sure this is the right direction.
I need more than just a solution, I need full explanation please
 A: Actually, you should notice that
$$f(x)=(x+1)\cos(x)$$
If you take the derivative at $x=0$, you end up with
$$f'(0)=\lim_{t\to0}\frac{(t+1)\cos(t)-1}t$$
A: I think it might be helpful to break the numerator up:
$$\lim_{t \to 0} \frac{(t+1)\cos t-1}{t}=\lim_{t \to 0} \frac{t\cos t+\cos t-1}{t}$$
Now, we have one term with $t\cos t$ and another part with $\cos t-1$. We know that as $t \to 0$, we have $\cos t \to 1$ and thus $\cos t-1 \to 0$. Therefore, let's break the limit up. First, do the $t\cos t$ part:
$$\lim_{t \to 0} \frac{t\cos t}{t}=\lim_{t \to 0} \cos t=1$$
Now, for the $\cos t-1$ part, we get $\cos t-1\to 0$ in the numerator and $t \to 0$ in the denominator, so we need to simplify further. Because we have $\cos t-1$ in the numerator, I am going to multiply by the "conjugate," which is $-\cos t-1$:
$$\frac{\cos t-1}{t}\frac{-\cos t-1}{-\cos t-1}=\frac{1-\cos^2 t}{t}\frac{1}{-\cos t-1}=\frac{\sin^2 t}{t}\frac{1}{-\cos t-1}=\frac{\sin t}{t}\frac{\sin t}{-\cos t-1}$$
It is well-known (from Squeeze Theorem) that $\lim_{t \to 0} \frac{\sin t}{t}=1$. Also, $\lim_{t \to 0} \frac{\sin t}{-\cos t-1}=\frac{\sin 0}{-\cos 0-1}=0$, so the whole second part is $1\cdot 0=0$.
This gives us the whole answer, first part plus second part, to be $1+0=1$.
A: Without calculus: 
Start with the geometrically obvious formula
$$\lim_{x\to 0} \, \frac{\sin (x)}{x}=1$$
Next use the triginometric identity
$$\cos (t)=\cos ^2\left(\frac{t}{2}\right)-\sin ^2\left(\frac{t}{2}\right)=1-2 \sin ^2\left(\frac{t}{2}\right)$$
Both relations together lead to the approximation 
$$\cos (t)\simeq 1-\frac{t^2}{2}$$
for $t\to 0$.
Inserting this into the expression in question and simplifying elementarily you can safely let $t\to 0$ to find the limit 1.
