AboutPaul Klarreich Expertise All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions.
I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.
Experience I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.
Question QUESTION: Well here's my situation. I haven't taken any math since high school and now I'm taking calculus. So i need the extra help. I have an exam on monday covering limits, which I do understand to an extent. But I'm stuck on these two problems. I'll write them down and then do as much as I can. Thanks for your help.
2x^2=2 lim x->2
1) 2x^2-2 < €
2) 2(x^2 -1) <€
3) 2(x-1)(x+1) <€
and now i'm stuck because the solution manual does this:
4) 6(x-1) < € if x-1< €/6, delta= min(1, €/6)
I don't understand where they get the 6 from, I'm guessing they plug in 2??? help on this part.
Next question:
Prove the limit is correct:
lim 1/(x+1)=-1 x->-2
My work:
1/(x+1) +1
1+x+1/x+1
x+2/x+1
now i'm lost from here on. Could you please help asap. thanks a lot!
ANSWER: Questioner: Gurvi
Category: Calculus
Private: No
Subject: Proving limits
Question: Well here's my situation. I haven't taken any math since high school and now I'm taking calculus. So i need the extra help. I have an exam on monday covering limits, which I do understand to an extent. But I'm stuck on these two problems. I'll write them down and then do as much as I can. Thanks for your help.
2x^2=2 lim x->2
1) 2x^2-2 < €
2) 2(x^2 -1) <€
3) 2(x-1)(x+1) <€
and now i'm stuck because the solution manual does this:
4) 6(x-1) < € if x-1< €/6, delta= min(1, €/6)
I don't understand where they get the 6 from, I'm guessing they plug in 2??? help on this part.
Next question:
Prove the limit is correct:
lim 1/(x+1)=-1 x->-2
My work:
1/(x+1) +1
1+x+1/x+1
x+2/x+1
now i'm lost from here on. Could you please help asap. thanks a lot!
.....................................................
Hi, Gurvi,
For starters, I will use e for epsilon. [I don't trust this site to get an epsilon through.]
Now whar you are writing is not so clear. Do you mean to prove that
Lim 2x^2 = 2 ??
x-> 1
If so, you want to prove that
|2x^2 - 2 | < e, when |x - 1| < d [delta]
2|x^2 - 1 | < e
2|x - 1||x + 1 | < e
Now you can make an assumption: if x is near 1, then x is between 0 and 2. In that case, |x + 1| is between 1 and 3. The worst case is that it's 3, and so if we can prove it for the worst case, we have it. So try to make:
2|x - 1|(3) < e
That's the same as:
6|x - 1| < e
and there is your 6. The rest is just:
|x - 1| < e/6,
so take d = e/6 and you can complete the proof.
1) 2x^2-2 < €
2) 2(x^2 -1) <€
3) 2(x-1)(x+1) <€
and now i'm stuck because the solution manual does this:
4) 6(x-1) < € if x-1< €/6, delta= min(1, €/6)
........................................
Next question:
Prove the limit is correct:
lim 1/(x+1)=-1 x->-2
1
lim ------ = -1
x->-2 x + 1
Again, you want to prove that
1
| ------ - (-1) | < e, when | x - (-2) | < d
x + 1
1
| ------ + 1 | < e
x + 1
1 + x + 1
| --------- | < e
x + 1
x + 2
| ------ | < e
x + 1
Now make a similar assumption: If x is near -2, it is between -3 and -1. Since x+1 is on the bottom, the WORST case is that it is small, not large. I.e. the worst case is x = -1. Oops -- that gives a zero on the bottom. Maybe we make it x between -3 and -1.5. That gives a worst case of x + 1 = -0.5:
| x + 2 |
| --------- | < e
| -1.5 + 1 |
| x + 2 |
| --------- | < e
| -0.5 |
| x + 2 |
| --------- | < e
1/2
| x + 2 | < 2e
And that's about it.
Yes, these are tricky.
---------- FOLLOW-UP ----------
QUESTION: I'm back again! okay i understand how you did these but quick question. In each one, how did you know that which one was the worst case?? I understand for problem 2 its not wise to use x=-1 since it will give a zero in the denominator but why do you choose -1.5 instead of -3 and the same goes for problem 1, why choose 3 instead of 1?? and
do you plug in the number you choose into the parenthesis you want to get rid of?? i mean which one do you stick the number into and why?? thanks again
Answer Hi, Gurvi,
You are using the
DEFINITION OF LIMIT:
lim f(x) = L if:
x->a
WHENEVER |x-a| is small, THEN |f(x) - L| is small.
How small? It means:
NEW DEFINITION OF LIMIT:
lim f(x) = L if:
x->a
Given an e, no matter how small, we can ALWAYS make |f(x) - L| < e, JUST by making |x-a| < some d, where the d will depend on the e.
So here's what you do: You write | f(x) - L | < e and you try to make it true. Somehow, you get your |x-a| in there. There will be other stuff. You will make the other stuff be as bad as it could be. Then you show that even then, you can still make your |f(x)-L| < e.
OK, your first: We got as far as:
2|x - 1||x + 1 | < e
There is your |x - 1|, which is the |x - a| part.
Now we want to make that true, provided |x-1| < d.
So solve the inequality:
e
| x - 1 | < ------- = delta
2|x+1|
Now we can make delta be that right side. But there is an x in there. So how small do we have to make it? The worst case would force us to choose a smaller delta. So how small? The worst case, making delta smaller, would have the bottom bigger. How big could it be?
Ah! That's where we make an assumption. In this case, we assume:
0 <= x <= 2, or
1 <= x+1 <= 3, or
2 <= 2|x+1| <= 6.
So the worst is that it's 6, making the smallest necessary delta equal to e/6.
Now you can run the proof:
Suppose |x-1| < e/6, our delta.
Then, since
6 >= 2|x+1|
1/6 <=1/2|x+1| << inverted
e/6 <= e/2|x+1| << multiply by e.
|x-1| < e/6 <= e/2|x+1| << transitive
|x-1| < e/2|x+1| << transitive
2|x+1||x-1| < e << multiply
2|x^2-1| < e << simplify
|2x^2-2| < e << simplify
|f(x) - L| < e << DONE.
And you called this a 'quick question.'
Don't worry -- the idea will come to you after a couple of years and a few hundred examples. That's how long it took me.
