# Homework Help: Showing by definition that limits exist

1. Sep 28, 2004

### physicsuser

I am having trouble with showing by definition that limits exist. For example limit as x goes to 3 for 5(x^2)-17

3-D < x < 3+D

28-E < 5x^2 - 17 < 28+E
45-E < 5x^2 < 45+E
9-E/5 < x^2 < 9+E/5
sqrRoot(9-E/5) < x < sqrRoot(9+E/5)

so does it mean that 3-D=sqrRoot(9-E/5) ?
then D=3+-sqrRoot(9-E/5)? If it is correct, can you do it without knowing the limit?

In linear equation you can solve it without even knowing the limit:

lim x goes to 3 for 5x - 17

3-D < x < 3+D

L=limit (5*3-17=-2)

L-E < 5x - 17 < L+E
L+17-E < 5x < L+17+E
(L+17)/5 - E/5 < x < (L+17)/5 + E/5

(L+17)/5=3
L+17=15
L=-2

3- E/5 < x < 3+ E/5

So D=E/5

2. Sep 28, 2004

### arildno

1. When doing limits exercises, introduce "epsilon" at the latest possible stage in your actual calculations!
2. Let's take your exercise in detail:
Let $$|x-3|<\delta$$:
Then, we have, guessing at 28 as our limit:
$$|(5x^{2}-17)-28|=5|x-3||x+3|$$
Assume $$\delta<1$$
Hence,
$$|(5x^{2}-17)-28|=5|x-3||x+3|<5\delta{4}=20\delta$$
By choosing $$\delta=minimum(1,\frac{\epsilon}{20})$$
we have bounded our error within a margin of $$\epsilon$$

3. Sep 28, 2004

### physicsuser

how do you get $$5|x-3||x+3|<5\delta{4}=20\delta$$ ?

4. Sep 28, 2004

### physicsuser

This is as far as I get.......

|x-3|<D
|5x^2-17 - 5x0^2+17|<E
5|x^2-x0^2|<E
|(x-x0)(x+x0)|<E/5

x0=3
and |x-3|<D
then D=E/(5|x+3|)

How do you get 20?

5. Sep 29, 2004

### arildno

By assuming D<1, |x+3|<4.
As I've said, do not introduce E in your calculations before you have to.

6. Sep 29, 2004

### physicsuser

I see, but this does not apply to every function does it?

lim as x goes to 0 for 1/x^2

|x-x0|<D

|1/x^2 - 1/x0^2| =
|(x0^2-x^2)/(x^2)(x0^2)| =
|(x^2-x0^2)/(-x0^2)(x^2)| =
|(x-x0)(x+x0)/(-x0^2)(x^2)|<E

D=E(-x0^2)(x^2)/(x+x0)
and since x0=0 then D=0 ???