Compute limit without L'Hospital's rule or derivative

Question

$\displaystyle\lim _{x\to 0}\left(\frac{2\cos x-2}{3x}\right)\:$

Answer 1

The first question amounts to finding $$A = \lim_{x \to 0} \frac{\cos x - 1}{x}$$ Replacing $x$ with $2x$ should yield the same limit: $$A = \lim_{2x \to 0} \frac{\cos 2x - 1}{2x} = \lim_{x \to 0} \frac{2\cos^2 x - 2}{2x} = \lim_{x \to 0} (\cos x + 1)\frac{\cos x - 1}{x} = 2A$$ so $A=0$.

For the second, the numerator and denominator of the fraction tend to $1$, so the answer is $1$.

Answer 2

Use the facts that:

$$\cos (2a) = 1 - 2 \sin^2 a \quad \text{(this is a well known formula)}$$

and $$\lim_{x \to 0} \frac{\sin mx}{nx} = \frac m n \quad \text{(an immeadiate consequence of $\lim_{x \to 0} \frac{\sin x}{x} =1$)}$$