I have to find this limit without using l'Hôspital's rule:
limx→0αsinβx−βsinαxx2sinαx
Using L'Hôspital's rule gives:
β6(α2−β2)
I am stuck where to begin without using the rule.
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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