Answer
I would consider limx→∞√x4−3x2−1−x2
as limx→∞√x4−3x2−1−x21
Then multiply √x4−3x2−1+x2 top and bottom to get rid of the ridiculous looking square root sign on the top. Then bingo!
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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