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Nikon285

09-19-2002, 01:20 PM

We just started a new section in Calculus today on the formal definition of limit and I'm quite confused on how to solve the problems. If anyone can show me how by using these two example problems to give me a better idea of what I'm doing, that would be great. I appreciate any help anyone can offer me.

Using the formal definintion of limit, prove:

1) lim (3x+2)=5

x->1

2) lim (1+x/2)=3/2

x->1

Thanks to anyone who can help me out,

Jason

minger

09-19-2002, 02:34 PM

I don't think that I ever learned anything about a formal defintion. Unless that's the one with the greek letter or whatever. Anyways, if you were to graph the function, f(x)=(3x+2) you can see that f(x)=5 as you approach 1 from both sides. Basically with a continuous function (a graph that can be drawn without lifting your pencil off of the paper) you can find the limit of a function by just plugging the limit into the equation.

It will get fun when your have things that are find the limit as the function approaches infinity.

muckyMuckMan

09-19-2002, 08:43 PM

This is probably a little to late but....

Using the formal definition of a limit (the epsilon-delta definition - look it up):

lim (3x+2)=5

x->1

Show that for all epsilon > 0, there exist delta > 0 such that |(3x+2) - 5| < epsilon when 0 < |x - 1| < delta.

In other words, find a way to make |(3x+2)-5| look like |x-1|.

First evaluate what's inside |(3x+2)-5|:

|3x-3| < epsilon

Factor out the three:

|3(x-1)| < epsilon

|3||x-1| < epsilon

Divide by |3|

|x-1| < epsilon/|3|

Since the absolute value of three is 3, you can take away the absolute value sign.

|x-1| < epsilon/3

Remember that 0 < |x-1| < delta from way above? See it? Okay. Since 0 < |x-1| < delta, and I just showed you that |x-1| < epsilon/3, we can choose delta = epsilon/3. This will give us

0 < |x-1| < epsilon/3

The point is to find a connection between epsilon and delta. That's the proof. You can do the other one doing the same thing I did here.

These things are a bit abstract. But I hope this helped a little.

minger

09-20-2002, 02:38 AM

yea, epsilon delta...that was it. man i hated that stuff.