Cfrgtkky

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

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.

A little more details:

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

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

hint

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

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

You will find $sqrt{2}$.

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