I wonder how to do this in different way from L'Hôpital's rule:
limx→02sinx−sin2xx−sinx.
Please help me solve this without using L'Hopital's 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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