My question is, how can I evaluate limits without L'Hopital's Rule ?
lim
Answer
For small x, 1-\cos x= 2\sin^2\frac{x}{2}\approx\frac{x^2}{2} while \sin\sqrt[3]{x}\approx\sqrt[3]{x}, so \frac{\sin\sqrt[3]{x}}{1-\cos x}\approx 2x^{-5/3} diverges as x\to 0. Thus the limit is \infty.
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