Evaluating trig limit $\lim\limits_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$

Question

Evaluate:
$$\lim_{x\to 0}\dfrac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$

I have tried to simplify the expression using the identity $1-\cos(x) = 2 \sin^2 (x/2)$, but I have still failed to remove the indeterminate form.

Answer 1

A little more details:

$$\lim_{x \to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)} = \lim_{x \to 0} \frac{\sqrt{2\sin^2(\frac{x^2}{2})}}{2 \sin^2 \frac{x}{2}} = \frac{\sqrt{2}}{2} \lim_{x \to 0} \frac{\sin(\frac{x^2}{2})}{\sin^2 \frac{x}{2}} = \sqrt{2}\lim_{x \to 0} (\frac{\sin(\frac{x^2}{2})}{\frac{x^2}{2}}\frac{\frac{x^2}{4}}{\sin^2\frac{x}{2}})$$

Answer 2

Hint: $$1-\cos\alpha=2\sin^2\dfrac{\alpha}{2}$$ Apply this with both numerator and denominator. Then use $$\lim_{x\to 0}\dfrac{\sin x}{x}=\lim_{x\to 0}\dfrac{x}{\sin x}=1$$

Answer 3

hint

$$\lim_{X\to 0}\frac{1-\cos(X)}{X^2}=\frac 12$$

and $$\sqrt{1-\cos(x^2)}=x^2\sqrt{\frac{1-\cos(x^2)}{(x^2)^2}}$$

You will find $\sqrt{2}$.

Answer 4

Multiply by conjugates: $$\lim_{x\to 0}\dfrac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\lim_{x\to 0}\dfrac{\sin(x^2)}{\sin^2(x)}\cdot \frac{1+\cos(x)}{\sqrt{1+\cos(x^2)}}=\\ \lim_{x\to 0}\dfrac{\sin(x^2)}{x^2}\cdot \dfrac{x^2}{\sin^2(x)}\cdot\frac{2}{\sqrt{2}}=\sqrt{2}.$$