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tntitansgirl · Jun 16, 2008, 08:48 PM

Use the Epsilon-Delta definition to prove the limit is L.

lim (4x+9)=21
x-> 3

galactus · Jun 17, 2008, 05:13 AM

$\lim_{x\to 3}(4x+9)=21$

We must show that given any positive number ${\epsilon}$ , we can find another

positive number ${\delta}$ such that $f(x)=4x+9$ satisifies

$|(4x+9)-21|<{\epsilon}$ ... [1]

Whenever x satisfies $0<|x-3|<{\delta}$ ... [2].

To find ${\delta}$ , we can rewrite [1] as $|4x-12|<{\epsilon}$

or $4|x-3|<{\epsilon}$

or $|x-3|<\frac{\epsilon}{4}$ ... [/3]

We must choose ${\delta}$ so that [3] is satisfied whenever [2]

is satisfied. So we take ${\delta}=\frac{\epsilon}{4}$ .

Now, can you see that the last step and finish?

tntitansgirl · Jun 17, 2008, 08:50 AM

Originally Posted by galactus

Now, can you see that the last step and finish?.

4(E/4) = E

Is that the last step? Thanks for your help! You are awesome, galactus!

galactus · Jun 17, 2008, 10:08 AM

To prove this choice works for ${\delta}$ , suppose that x satisfies [2].

Since we are letting ${\delta}=\frac{\epsilon}{4}$ , it follows from [2]

that x satisfies $0<|x-3|<\frac{\epsilon}{4}$

Therefore, [3] is satisfied since it is just the right hand inequality in $0<|x-3|<\frac{\epsilon}{4}$

This proves that $\lim_{x\to 3}(4x+9)=21$