Question

prove the statement lim x-> 1 (2 + 4x) / 3 = 2

Answer 1

\[\text{ solve } |f(x)-L|<\epsilon \text{ for } |x-a| \]
where a is the number x approaches

Answer 2

for example pretend I have
lim x->3 (5-3x)=5-3(3)=5-9=-4
|f(x)-L|<e
|5-3x-(-4)|<e
|5-3x+4|<e
|-3x+9|<e
remember we are trying to solve this for |x-3|
|-3*(x-3)|<e
|-3||x-3|<e
3|x-3|<e
|x-3|<e/3
so you will choose delta to be e/3 for this example

Answer 3

so do you think you can solve |(2+4x)/3-2|<e for |x-1| ?

Answer 4

So far I got |x-3|<3/4e

Answer 5

should be |x-1|<3/4 *e

Answer 6

\[|\frac{2+4x}{3}-2|<\epsilon \\ |\frac{2+4x}{3}-\frac{6}{3}|< \epsilon \\ |\frac{2+4x-6}{3}|< \epsilon \\ |\frac{4x-4}{3}|<\epsilon \\ \frac{4}{3}|x-1|<\epsilon \\ |x-1|<\frac{3}{4} \epsilon \]
this means you will begin your proof with something like
Let epsilon>0, choose delta=3 epsilon/4 .... Then just just |f(x)-L|<epislon with the delta you found

Answer 7

oops that's typo. I DID have |x-1|