• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Quick calculus question (i think)

I

iman00b

Member
Anyone think they can help me with random problems i get stuck on? Only help, not doing all the work for me.

prove lim x->4 (7-3x) = -5 using (epsilon delta) definition. E = Epsilon D = Delta

|f(x) - L| < E whenever 0<|x-a|<D
|(7-3x) - (-5)| < E whenever 0<|x-4|<D
|12-3x|<E ...
-3|x-4|<E ...
|x-4|> -E/3 ...

The < sign switches to > because im dividing by a negative correct? If so, |x-4|<D and |x-4|> -E/3 dont match up so i cant find what D equals. My teacher is completely useless so im trying to follow the book. All i understand right now is to break down |f(x) - L| < E to look like 0<|x-4|<D, to find out what D equals. Then use 0<|x-4|<D and expand(?) to make it look like |f(x) - L| < E again and theres the proof.
 
I actually kinda like math, i just hate it when im stuck. Its like a puzzle i cant figure out. English I hate, since theres no one answer to always get it right.
 
Unless you're a math major, you'll never really have to do a problem like that again.

Sorry, can't help you, I hardly knew how to do it when I "learned" it 5 years ago.
 
-3|x-4|<E ...
|x-4|> -E/3 ...

The sign is supposed to change though right? If its not then
|x-4|< -E/3 whenever 0<|x-4|<D
D = -E/3 and i could finish solving the problem. But i dont think its right.
 
you can break down the absolute value equation into 2 separate ones, dont remember exactly

Yes true, but not needed. If |x-4|<D and |x-4|< -E/3, you can conclude that D = -E/3 eliminating the need to split it into two. Well, thats my understanding as of now. If i do split it into two, i dont know what i would do in the next step since none of the examples show it splitting into two.
 
Originally posted by: iman00b
you can break down the absolute value equation into 2 separate ones, dont remember exactly

Yes true, but not needed. If |x-4|<D and |x-4|< -E/3, you can conclude that D = -E/3 eliminating the need to split it into two. Well, thats my understanding as of now. If i do split it into two, i dont know what i would do in the next step since none of the examples show it splitting into two.
Click to expand...

you cant conclude that they are equal, one can be larger than the other, and yet the first value can still be larger than |x-4|
 
Originally posted by: iman00b

|12-3x|<E ...
-3|x-4|<E ...
Click to expand...

I am pretty sure the sign is not supposed to change there. For example, |-3x| = |3x| = 3|x|.

|12-3x|<E
|(-1)(3)(x-4)|<E
|(3)(x-4)|<E
3|x-4|<E
...
 
Originally posted by: MSCoder610
Originally posted by: iman00b

|12-3x|<E ...
-3|x-4|<E ...
Click to expand...

I am pretty sure the sign is not supposed to change there. For example, |-3x| = |3x| = 3|x|.

|12-3x|<E
|(-1)(3)(x-4)|<E
|(3)(x-4)|<E
3|x-4|<E
...
Click to expand...

ding ding ding!
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top