lim x→0 √(1 - cos(x²)) / (1 - cos x)

Explanation:

Step1: Use the double - angle formula

Recall that $1 - \cos\alpha=2\sin^{2}\frac{\alpha}{2}$. So, $1 - \cos(x^{2}) = 2\sin^{2}\frac{x^{2}}{2}$ and $1 - \cos x=2\sin^{2}\frac{x}{2}$. The limit becomes $\lim_{x\rightarrow0}\frac{\sqrt{2\sin^{2}\frac{x^{2}}{2}}}{2\sin^{2}\frac{x}{2}}$. Since $x\rightarrow0$, $\sin\frac{x^{2}}{2}\geq0$ and $\sin\frac{x}{2}\geq0$, so it is $\lim_{x\rightarrow0}\frac{\sqrt{2}\sin\frac{x^{2}}{2}}{2\sin^{2}\frac{x}{2}}$.

Step2: Use the small - angle approximation

When $t\rightarrow0$, $\sin t\sim t$. As $x\rightarrow0$, $\sin\frac{x^{2}}{2}\sim\frac{x^{2}}{2}$ and $\sin\frac{x}{2}\sim\frac{x}{2}$. Substitute these into the limit: $\lim_{x\rightarrow0}\frac{\sqrt{2}\cdot\frac{x^{2}}{2}}{2\cdot(\frac{x}{2})^{2}}$.

Step3: Simplify the expression

$\lim_{x\rightarrow0}\frac{\sqrt{2}\cdot\frac{x^{2}}{2}}{2\cdot\frac{x^{2}}{4}}=\lim_{x\rightarrow0}\frac{\sqrt{2}\cdot\frac{x^{2}}{2}}{\frac{x^{2}}{2}}=\sqrt{2}$.

Answer:

$\sqrt{2}$