So I made one exercise, which was limx→+∞1√x2−2x−x I solved this one by:
limx→+∞1√x2−2x−x√x2−2x+x√x2−2x+x
limx→+∞√x2√1−2x+x−2x
limx→+∞√1−2x+1−2=−1
The next exercise is limx→−∞1√x2+2x−x
I thought that I could solve this one in the same way by:
limx→−∞1√x2+2x−x√x2+2x+x√x2+2x+x
limx→−∞√x2√1+2x+x2x
limx→−∞√1+2x+12=1
But apparently, the answer is 0...
Why can't the second one be solved in the same way as the first?
And why is the answer 0?
Answer
When x→−∞, you don't have 0 as denominator, but +∞, because when x→−∞,√x2−2x∼|x| and, as x<0, |x|=−x, so you have limx→−∞1−x−x=limx→−∞1−2x=0.
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