• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the largest value of delta greater than 0

  • Thread starter zurburgk
  • Start date
  • #1
2
0

Homework Statement


For f(x) = sqrt(27-x), L=4, x_0 = 11 epsilon = 1, find the largest value of delta > 0 in the formal definition of a limit which ensures that |f(x) - L| < epsilon


Homework Equations



the formal def. of a limit:
lim x->x_0 F(x) = L if, for every number epsilon > 0, there exists a corresponding number delta>0 such that
0<|x-x_0| < delta implies |f(x) - L| < epsilon


The Attempt at a Solution



I've tried plugging multiple different things in to the equations, but i, sadly, have no idea what i'm doing.

0 < |x-11| < epsilon and |sqrt(27-x) - 4| < 1.

Attempting to solve the second equation i get x>2. If i plug that into the first equation i got -9. But that is definitely not the answer. I just don't know what all these numbers mean.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998

Homework Statement


For f(x) = sqrt(27-x), L=4, x_0 = 11 epsilon = 1, find the largest value of delta > 0 in the formal definition of a limit which ensures that |f(x) - L| < epsilon


Homework Equations



the formal def. of a limit:
lim x->x_0 F(x) = L if, for every number epsilon > 0, there exists a corresponding number delta>0 such that
0<|x-x_0| < delta implies |f(x) - L| < epsilon


The Attempt at a Solution



I've tried plugging multiple different things in to the equations, but i, sadly, have no idea what i'm doing.

0 < |x-11| < epsilon and |sqrt(27-x) - 4| < 1.

Attempting to solve the second equation i get x>2. If i plug that into the first equation i got -9. But that is definitely not the answer. I just don't know what all these numbers mean.
How about solving [itex]\displaystyle |\sqrt{27-x\ } - 4| = 1[/itex] for x, to start things off ?
 
  • #3
2
0
Do you mind if i walk you through my steps, if i make a mistake it should be easier to pinpoint the area :)

So |sqrt(27-x)-4| < 1
- add 4 to both sides

|sqrt(27-x)| < 5
-get rid of the square root

|27-x|< 25
-subtract 27 from each side

|-x| < -2

- and here is where i'm having my first problem. Do i multiply both sides by a negative allowing x to be a positive variable? Or do the absolute value signs take care of the negative variable?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998
Do you mind if i walk you through my steps, if i make a mistake it should be easier to pinpoint the area :)

So |sqrt(27-x)-4| < 1
- add 4 to both sides

|sqrt(27-x)| < 5
-get rid of the square root

|27-x|< 25
-subtract 27 from each side

|-x| < -2

- and here is where i'm having my first problem. Do i multiply both sides by a negative allowing x to be a positive variable? Or do the absolute value signs take care of the negative variable?
Let me repeat,
Solve [itex]\displaystyle |\sqrt{27-x\ } - 4| = 1[/itex] for x.​

But if you insist, you can solve the inequality. Just be careful. That is trickier.

At any rate:
[itex]\displaystyle |\sqrt{27-x\ } - 4|+4\ne |\sqrt{27-x\ }|[/itex]​

If [itex]\displaystyle |\sqrt{27-x\ } - 4| < 1[/itex]

then [itex]\displaystyle -1<\sqrt{27-x\ } - 4 < 1\ .[/itex]

Can you continue from there?
 

Related Threads on Finding the largest value of delta greater than 0

Replies
3
Views
1K
  • Last Post
Replies
3
Views
919
Replies
10
Views
769
Replies
8
Views
762
Replies
16
Views
3K
  • Last Post
Replies
3
Views
3K
Replies
6
Views
2K
Replies
3
Views
11K
Replies
2
Views
3K
Replies
1
Views
6K
Top