# Prove A Left Hand Limit

1. Aug 31, 2010

### TrueStar

1. The problem statement, all variables and given/known data

Proove the statement using the $$\epsilon$$ and $$\delta$$ definition of a limit.

The limit of x as it approaches 9 from the left side of the x axis is $$\sqrt[4]{9-x}$$=0

2. Relevant equations

The limit of x as it approaches a from the left side of the x axis is f(x) = L

if for every number $$\epsilon$$>0 there is a number $$\delta$$>0 such that if

a-$$\delta$$<x<a then |f(x)-L|<$$\epsilon$$

3. The attempt at a solution

9-$$\delta$$<x<9 and |$$\sqrt[4]{9-x}$$-0|<$$\epsilon$$

Using algebra, I've worked my way to |9-x|<$$\epsilon^{4}$$. I'm not sure where to go with it now. Do I subtract the 9 over and then change the sign of x, ending up with |x|>-$$\epsilon$$$$^{4}$$+9 ? That doesn't seem right to me.

2. Aug 31, 2010

### ╔(σ_σ)╝

You can pick delta=episilon^4 and you are done.

3. Aug 31, 2010

### TrueStar

That's great! However, I don't know why it's OK to stop there. Would anyone be kind enough to explain?

More specifically, why can I stop at |9-x| and how do I work my way back to verify that $$\epsilon$$$$^{4}$$ can be substituted for $$\delta$$?

4. Aug 31, 2010

### ╔(σ_σ)╝

You can stop there because this is the precise definitionof a limit. We say limit as x approaches a f(x)=L if
When|f(x)-L|< episilon then |x-a|< delta.

If your case a=7 and delta = episilon^4.

Too lazy to use latex of my phone, sorry!

5. Aug 31, 2010

### zooxanthellae

I believe it's because if we assume that d = E^4, then we have 9 - E^4 < x < 9. This means that 9 - E^4 - x < 0 < 9 - x. Then 9 - x < E^4 < 9 - x + E^4.

Then you have the algebra: since |(9 - x)^(1/4)| < E, then |9 - x| < E^4 in order for the limit to be true. We've found a delta such that this is true, so our proof is complete.

6. Aug 31, 2010

### TrueStar

Thank you both so much for explaining it for me! I understand why it works now. There were no good examples in the textbook, so I can use this as a guide for similar problems.