My question is, how can I evaluate limits without L'Hopital's Rule ?
limx→0sin(3√x)1−cosx
Answer
For small x, 1−cosx=2sin2x2≈x22 while sin3√x≈3√x, so sin3√x1−cosx≈2x−5/3 diverges as x→0. Thus the limit is ∞.
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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