Is my proof of $\lim_{x\rightarrow 0} \frac{\cos(x) - 1}{x} = 0$ correct?

Question

Let $$\lim_{x\rightarrow 0} \frac{\cos(x) - 1}{x} = h$$ then $$\lim_{x\rightarrow 0}h^2 = \frac{\cos^2(x) - 2\cos(x) + 1}{x^2}$$ substituting $$\cos^2x = 1 - \sin^2x$$ gives $$\lim_{x\rightarrow 0}h^2 = \frac{2 - 2 \cos(x) - \sin^2x}{x^2}$$Since $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 $$ It squared must also equal $1$. So $$ h^2 = \lim_{x\rightarrow 0} \frac{2 - 2 \cos(x)}{x^2} - 1 = \lim_{x\rightarrow 0} 2\left(\frac{1-\cos(x) }{x}\right) . \frac{1}{x} - 1 = \lim_{x\rightarrow 0} \frac{2h}{x} - 1$$

If $h$ was a nonzero limit then the numerator would converge and since $\lim_{x\rightarrow 0}\frac{1}{x} = \infty$ the fraction would diverge to infinity. But since the limit cannot be infinity then the limit must be $0$.

Answer 1

You are assuming from the beginning that the limit exist, and you apply algebra of limits without knowing if it converges, so is not correct altough the answer is correct

Answer 2

I would agree with the comments above that you're assuming it already exists. You should try to approach it by manipulating only the limit. So, I liked your logic of using the Pythagorean identity. Think about multiplying both the numerator and denominator by an expression that will give you $1-cos^2(\theta)$ or similar to it $cos^2(\theta)-1=-sin^2(\theta)$).