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

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

\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?

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!

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