Wednesday, 12 November 2014

How can I find limits without L'Hopital's Rule?





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.


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...