The limit is
limx→π/41−tanx1−√2sinx
I was able to solve it using L'hopital and the answer that I got was 2.
Can you please confirm if the answer is right and suggest some other way to evaluate the limit without using L'hopital?
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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