Proving Limits at Infinity: Can You Help Me with These Two Limits?

In summary, the conversation discusses how to prove two limits using the definition of limit. The first limit is \lim_{x\to\infty}\frac{x-1}{x+2} = 1, and the second limit is \lim_{x\to-1}\frac{-1}{(x+1)^2} = -\infty. The steps for proving these limits are explained, with a focus on simplifying expressions and finding appropriate values for $N$ or $\delta$ in order to satisfy the given conditions.
  • #1
goody1
16
0
Hi, can anybody help me with this two limits? I have to prove them by the definition of limit. Thank you in advance.
View attachment 9630 View attachment 9631
 

Attachments

  • limitka.png
    limitka.png
    1.1 KB · Views: 61
  • limitka2.png
    limitka2.png
    1.3 KB · Views: 63
Mathematics news on Phys.org
  • #2
goody said:
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.
 
  • #3
Hi Opalg! Do you think I got it correct?
View attachment 9633
 

Attachments

  • what.png
    what.png
    2.8 KB · Views: 58
  • #4
goody said:
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.
 

1. What is the definition of a limit?

The limit of a function is the value that a function approaches as the input value gets closer and closer to a specific value. It is denoted by the notation lim f(x) as x approaches a.

2. How do you prove a limit by definition?

To prove a limit by definition, you must show that for any positive number ε (epsilon), there exists a corresponding positive number δ (delta) such that if the distance between the input value x and the limit value a is less than δ, then the distance between the function value f(x) and the limit L is less than ε.

3. What is the importance of proving a limit by definition?

Proving a limit by definition is important because it is the most rigorous way to establish the existence of a limit. It also allows us to understand the behavior of a function near a specific point and make predictions about its behavior.

4. What are some common techniques used to prove a limit by definition?

Some common techniques used to prove a limit by definition include algebraic manipulation, the squeeze theorem, and the epsilon-delta definition of a limit.

5. Can a limit be proven by definition if the function is not continuous?

Yes, a limit can still be proven by definition even if the function is not continuous. The epsilon-delta definition of a limit does not require the function to be continuous, only that it approaches a specific value as the input value gets closer and closer to a certain value.

Similar threads

Replies
11
Views
1K
Replies
14
Views
1K
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
780
Replies
16
Views
2K
Replies
7
Views
563
  • Calculus and Beyond Homework Help
Replies
5
Views
891
Replies
8
Views
2K
Replies
17
Views
2K
Replies
6
Views
1K
Back
Top