I used u substitution for the limit limx→∞x2sin1x
and got the limit does not exist by saying u=1x. Is this correct and if so would that mean limx→∞x3sin1x2 and limx→∞x3sin1x also don't exist?
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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