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- Feb 7, 2012

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Hi goody , and welcome to MHB!Hi, can anybody help me with this two limits? I have to prove them by the definition of limit. Thank you in advance.

To prove that \(\displaystyle \lim_{x\to\infty}\frac{x-1}{x+2} = 1\), you have to show that, given $\varepsilon > 0$, you can find $N$ such that \(\displaystyle \left|\frac{x-1}{x+2} - 1\right| < \varepsilon\) whenever $x>N$.

So, first you should simplify \(\displaystyle \left|\frac{x-1}{x+2} - 1\right|\). Then you should see how large $x$ has to be in order to make that expression less than $\varepsilon$.

To prove that \(\displaystyle \lim_{x\to-1}\frac{-1}{(x+1)^2} = -\infty\), you have to show that, given $M$, you can find $\delta>0$ such that \(\displaystyle \frac{-1}{(x+1)^2} < -M\) whenever $|x+1| < \delta$. That is actually an easier calculation than the first one, so you might want to try that one first.

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Not quite, although you started correctly. The limit in this case is as $x\to\infty$, so you want to see what happens when $x$ gets large. This means that the inequality $\dfrac3{|x+2|}<\varepsilon$ has to hold for all $x$ greater than $N$ (where you think of $N$ as being a large number).HiOpalg! Do you think I got it correct?

Write the inequality as $|x+2| > \dfrac3\varepsilon$, and you see that this will be true if $x > \dfrac3\varepsilon -2$. So you can take $N = \dfrac3\varepsilon -2$. More simply, you could take $N = \dfrac3\varepsilon$, which will satisfy the required condition with a bit to spare.