Use expansions: $\sin x = x - \frac{x^{3}}{6} + \cdots$ $\cos x = 1 - \frac{x^{2}}{2} + \cdots$ $\sqrt{x}$ near $0$ goes to $0$ Numerator: $ x + (x - \frac{x^{3}}{6}) = 2x + O(x^{3}) $ Denominator: $ \sqrt{x} - \cos x = \sqrt{x} - 1 + \frac{x^{2}}{2} + \cdots $ As $x\to 0$, denominator → $-1$ So limit = $0$.