Question

What am I doing wrong?
http://prntscr.com/caw3mf

Answer 1

My work:

Answer 2

And doing a similar problem with the same method, I got it correct:
http://prntscr.com/caw44j

Also, this is the \(\epsilon\)-\(\delta\) definition of a limit type of question :p

Answer 3

And the way the book solved that example which I got correct:
http://prntscr.com/caw4rn

They just used a different range or numbers, but my professor said that it doesn't matter which range you use. :P

Answer 4

since lightshot keeps going down:

Answer 5

and the similar example I did correctly with how they solved it

Answer 6

The \(\epsilon\)-\(\delta\) definition of a limit... just writing it for fun ;p

\[\lim_{x \rightarrow a} f(x) = L\]
\[\forall~\epsilon>0 ~~\exists~\delta>0 \]Then
\[0<|x-a|<\delta \implies |f(x)-L|<\epsilon \]

Answer 7

Haven't done this method in ages, never liked it >.<, but looks like you did it right? What exactly do you think you did wrong, some of the screen shots aren't working for me so I can't see

Answer 8

I feel like I did everything correct. And since lightshot kept going down, I uploaded the images also

Answer 9

Ok so we want \[| x^2+4| < \epsilon ~~~ if ~~~ |x-5| < \delta\] yeah?

Answer 10

Essentially what I'm saying is this \[|f(x)-c|< \epsilon ~~~ if ~~~ |x-a|< \delta\]

Answer 11

x^2 + 4 - 29

since L = 29

Answer 12

Right so your \[\epsilon = 0.01\] here correct?

Answer 13

Yes

Answer 14

Ok so we have then \[|x^2-4-29|< \epsilon ~~~ if~~~|x-5|< \delta\] I see this is what it should be since you have L yeah ok haha then lets carry on...we get \[|x^2-33| < \epsilon \] ooh dirty I don't like these numbers

Answer 15

Heh ok so something is wrong here

Answer 16

Oh stupid me I see it's x^2+4!!

Answer 17

yeah, it'll factor down to |x-5||x+5| < \(\epsilon\)

mhmm

Answer 18

\[|x^2+4-29|< \epsilon ~~~ if~~~|x-5|< \delta\]
then we have \[|x^2-25|< \epsilon \implies |(x-5)(x+5)|< \epsilon \] so we have \[|x-5| < \frac{ \epsilon }{ |x+5| }\] now is where it gets tricky we'll have to make a few assumptions

Answer 19

I think the problem is where you say 1 < x < 3
that should be 4<x<6 (in other words, the interval must contain 5, the value you are approaching)

Answer 20

\[\delta = \frac{ \epsilon }{ 11 } = \frac{ 0.01 }{ 11 } \approx 0.000909091\] heh

Answer 21

Yes I did the same thing as I assumed \[4 < x < 6\]

Answer 22

yes, astro has it.

Answer 23

***
They just used a different range or numbers, but my professor said that it doesn't matter which range you use. :P
***

Yes, that is true, but whatever interval you pick, it must contain the 5
if you use a large interval, I think you get a smaller delta (smaller than it has to be)
and a small interval lets delta be larger (relatively speaking)

Answer 24

Oh, so the range we choose must have c in it? mhmm

I'll keep that in mind from now on ^-^

And that was correct. Thanks all! :)

Answer 25

You probably don't need this, but fyi, these "moldy oldies" are pretty good:
http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-4-derivatives-and-limits/
http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-5-a-more-rigorous-approach-to-limits/

Answer 26

I'll take a look at them. Thanks astro and phi! I appreciate it! :)