need help with limit, just a refresher

**ffxwolf** · September 22nd 2009, 10:58 AM

Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3

**redsoxfan325** · September 22nd 2009, 11:28 AM

Originally Posted by ffxwolf

Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3

You want to find $\delta$ such that $|x-3|<\delta \implies |(4x-1)-11|<\epsilon$ .

Choose $\delta=\frac{\epsilon}{4}$ .

Thus when $|x-3|<\delta$ , we have that $|(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon$

**ffxwolf** · September 22nd 2009, 11:43 AM

Thank you for your response, I just don't understand why delta = epsilon/4

**redsoxfan325** · September 22nd 2009, 11:53 AM

Originally Posted by ffxwolf

Thank you for your response, I just don't understand why delta = epsilon/4

If you don't know what $\delta$ is, you can solve for it.

Whatever $\delta$ is you know that $|x-3|<\delta$ .

So, work with what you have. You ultimately want $|(4x-1)-11|<\epsilon$ .

So, similar to before,

$|(4x-1)-11|=|4x-12|=4|x-3|<4\delta$ .

Now you want $4\delta$ to be equal to $\epsilon$ , so you just solve for it.

$4\delta=\epsilon\implies \delta=\frac{\epsilon}{4}$

**ffxwolf** · September 22nd 2009, 11:58 AM

Okay, I see it now. Thanks

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