Evaluate $$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
My attempt: $\cos(\frac{1}{1-x})=\cos(\frac{1}{x-1})$, so if we let $t=x-1$ we get $$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{x-1})}{\sqrt{x-1}}=\lim_{x \to 0^+} \frac{\sin{((t+1)^3-1)}\cos(\frac{1}{t})}{\sqrt{t}}$$ That's where I got stuck because this doesn't look any better I guess. Any hint?
For $x\rightarrow1^+$ we obtain: $$\frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}=\frac{\sin(x^3-1)}{x^3-1}\cdot\sqrt{x-1}(x^2+x+1)\cos\frac{1}{x-1}\rightarrow0.$$