[SOLVED] Epsilon Delta (limit definition)

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January 20th 2010, 06:09 PM
Noxide

[SOLVED] Epsilon Delta (limit definition)

I'm given an epsilon value of 0.5
The limit of my function evaluated as x-->0 is 1

my function is

f(x) = (e^x - 1) / (x)

Just a restatement of the variable
e = 0.5
L = 1
f(x) = (e^x - 1) / (x)
x = ?
d(delta) = ?
a = 0

I've plugged the values into the inequalities and i'm stuck on this part:

|f(x)-L| < e
i.e.
| [(e^x - 1) / (x)] - 1| < 0.5
So, I simplify and then I get to this point:
e^x - 1 - 1.5x < 0
I can't figure out x and because I can't figure out x I'm unable to figure out delta. I need some help solving for x. I'm aware that I will likely need to apply a natural logarithm but whenever I do I still am unable to solve for x.
January 20th 2010, 07:37 PM
Jhevon

Quote:

Originally Posted by Noxide

I'm given an epsilon value of 0.5
The limit of my function evaluated as x-->0 is 1

my function is

f(x) = (e^x - 1) / (x)

Just a restatement of the variable
e = 0.5
L = 1
f(x) = (e^x - 1) / (x)
x = ?
d(delta) = ?
a = 0

I've plugged the values into the inequalities and i'm stuck on this part:

|f(x)-L| < e
i.e.
| [(e^x - 1) / (x)] - 1| < 0.5
So, I simplify and then I get to this point:
e^x - 1 - 1.5x < 0
I can't figure out x and because I can't figure out x I'm unable to figure out delta. I need some help solving for x. I'm aware that I will likely need to apply a natural logarithm but whenever I do I still am unable to solve for x.

Note that we want $|x - 0| = |x| < \delta$ . Lets say we want $\delta$ to be at most 1. so $|x| < 1 \implies -1 < x < 1 \implies e^{-1} < e^x < e$ .

Now, go back to our expression and apply this bound on $e^x$ :

$\bigg| \frac {e^x - 1}x - 1 \bigg| < \bigg| \frac {e - 1}x - 1 \bigg| < 0.5$

no solve the last inequality so that $x$ is bounded between two constants. $\delta$ will be the larger one
January 21st 2010, 08:39 AM
Noxide

Thanks again, I really appreciate your help.