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Proving limit by definition

goody

New member
Mar 31, 2020
14
Hi, can anybody help me with this two limits? I have to prove them by the definition of limit. Thank you in advance.
limitka.png limitka2.png
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,676
Hi, can anybody help me with this two limits? I have to prove them by the definition of limit. Thank you in advance.
Hi goody , and welcome to MHB!

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.
 

goody

New member
Mar 31, 2020
14
Hi Opalg! Do you think I got it correct?
what.png
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,676
Hi Opalg! Do you think I got it correct?
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).

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.