Question

Prove that lim as x-->3 (2x-6=0). (using the rigorous definition of limit)

Answer 1

\[\lim_{x \rightarrow 3}2x-6=0\]

Answer 2

you mean the epsilon-delta definition?

Answer 3

I'm not sure. I just plugged in 3 for x to show that it equals 0 but I got it wrong.

Answer 4

check this out https://www.khanacademy.org/math/differential-calculus/limits_topic/epsilon_delta/v/epsilon-delta-limit-definition-1

Answer 5

\[\text{ we want to show whenever } |x-a|<\delta \text{ then } |f(x)-L|< \epsilon \text{ if given } \\ \lim_{x \rightarrow a}f(x)=L\]

Answer 6

so go ahead and plug in your L, f(x) and your a

Answer 7

we are going to take |f(x)-L|<epsilon
and try to solve for |x-a|

Answer 8

this is not the proof
this is just us trying to find a delta that works

Answer 9

For example:
\[\lim_{x \rightarrow 5}(2x+5)=15\]
to prove this we must choose a delta such that whenever |x-a|<delta we then have |f(x)-L|<epsilon
We want to show the then part eventually using the if part
but first we must work backwards to find delta

\[|f(x)-L|<\epsilon \\ |2x+5-15|<\epsilon \\ |2x-10|<\epsilon\]
now remember we are trying to solve |f(x)-L|<epsilon for |x-a|
so we can choose are delta
you know that |x-a|<delta
so anyways
\[|2x-10|<\epsilon \\ |2| \cdot |x-5|<\epsilon \\ 2|x-5|<\epsilon \\ |x-5|< \frac{\epsilon}{2}\]
so we will want to choose delta=epsilon/2 for our proof
then for the actual proof part you will take this choosing of delta=epsilon/2 and show |f(x)-L|<epsilon