So I made one exercise, which was limx→+∞1√x2−2x−x I solved this one by:
\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}
\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}
\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1
The next exercise is \lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}
I thought that I could solve this one in the same way by:
\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x}
\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}
\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 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\rightarrow-\infty, you don't have 0 as denominator, but +\infty, because when x \rightarrow-\infty, \sqrt{x^2 -2x} \sim |x| and, as x < 0, |x| = -x, so you have \lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0.
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