Finding the largest value of delta greater than 0

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To find the largest delta greater than 0 for the limit problem involving f(x) = sqrt(27-x), L=4, and x_0=11 with epsilon=1, the key is to solve the inequality |sqrt(27-x) - 4| < 1. This leads to the expression -1 < sqrt(27-x) - 4 < 1, which simplifies to 3 < sqrt(27-x) < 5. Squaring these inequalities results in 9 < 27 - x < 25, giving the bounds for x as 2 < x < 18. The largest delta is then determined by the distance from x_0=11 to the nearest endpoint, which is 2, resulting in delta = 9. Thus, the largest value of delta that satisfies the condition is 9.
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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 into 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.
 
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zurburgk said:

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 into 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 \displaystyle |\sqrt{27-x\ } - 4| = 1 for x, to start things off ?
 
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?
 
zurburgk said:
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 \displaystyle |\sqrt{27-x\ } - 4| = 1 for x.​

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

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

If \displaystyle |\sqrt{27-x\ } - 4| &lt; 1

then \displaystyle -1&lt;\sqrt{27-x\ } - 4 &lt; 1\ .

Can you continue from there?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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