Question

I need to prove the limit statement lim x-> (9-x)=5

Answer 1

x approaches 4 right

Answer 2

yes approaches 4

Answer 3

\[|f(x)-L|<\epsilon \\ |(9-x)-5|<\epsilon \text{ solve this inequality for } |x-4|\]

Answer 4

\[\text{ EXAMPLE } \\ \text{ Prove } \lim_{x \rightarrow 3}(4x-1)=11 \\ \text{ so we want } \\ |x-a|<\delta \text{ whenever } |f(x)-L|<\epsilon \\ \\ \text{ For my example } \\ a=3 \\ f(x)=4x-1 \\ L=11 \\ \text{ so we need \to solve } |4x-1-11|<\epsilon \text{ for } |x-3| \\ |4x-1-11|<\epsilon \\ |4x-12|<\epsilon \\ |4(x-3)|<\epsilon \\ |4| \cdot |x-3| <\epsilon \\ 4 |x-3| <\epsilon \\ |x-3|<\frac{\epsilon }{4} \\ \text{ So we will need \to choose } \delta=\frac{\epsilon }{4} \]
So this above was just finding delta.
We need to prove this delta works.
\[\text{ Given } \epsilon>0, \text{ let } \delta=\frac{\epsilon}{4}. \text{ So we have for whenever } |x-3|<\delta , \\ |f(x)-L|=|(4x-1)-11|= |4x-12|=4|x-3|<4 \cdot \delta =4 \cdot \frac{\epsilon}{4} =\epsilon \\ \text{ So we have proved by our definition of limit that } \\ \lim_{x \rightarrow 3} (4x-1)=11 \]

Answer 5

the bottom 4 lines is actually the proof

Answer 6

\[|9-x-5|<\epsilon \text{ so we need \to solve this for } |x-4| \\ |-1(x-4)|<\epsilon \\ |-1| |x-4|<\epsilon \\ 1 |x-4|<\epsilon \\ |x-4|<\epsilon \]
so here you would choose delta=epsilon

Answer 7

ok is see the problem i was having, I had been separating \[\left| 9-x-5 \right| as (9-x)-5\]

Answer 8

basically we are trying to find some delta such that |f(x)-L|<epislon

Answer 9

where delta is a function of epsilon

Answer 10

so you were dropping the | |?

Answer 11

no I was adding the -5 from epsilon on the right side

Answer 12

oh sounds dangerous

Answer 13

mainly because the -5 is inside with the rest of the stuff like you said :)

Answer 14

\[\text{ in general for lines only! other functions require more creativity } \\ \text{ Say we have } \\ \lim_{x \rightarrow a } (bx+c)=ba+c \\ \text{ First let's find } \delta \\ |f(x)-L|<\epsilon \text{ solve for } |x-a| \\ |(bx+c)-(ba+c)|<\epsilon \\ |b(x-a)|<\epsilon \\ |b| |x-a|<\epsilon \\ |x-a|<\frac{\epsilon}{|b|} \\ \text{ So this means we will choose } \\ \delta=\frac{\epsilon}{|b|}\]
Now for the proof:
\[\text{ Given } \epsilon>0, \text{ let } \delta=\frac{\epsilon}{|b|}. \text{ So we have for whenever } |x-a|<\delta , \\ |f(x)-L|=|(bx+c)-(ba+c)|= |b(x-a)|=|b||x-a|<|b| \cdot \delta =|b| \cdot \frac{\epsilon}{|b|} =\epsilon \\ \text{ So we have proved by our definition of limit that } \\ \lim_{x \rightarrow a} (bx+c)=ba+c\]

Answer 15

line got cut off basically i have found a delta in terms of epsilon such that |f(x)-L|<epsilon

Answer 16

I am following what your saying, cleared things up nicely, thank you.

Answer 17

np