Finding the range of (1-(1/x))^x from x=1 to infinite

In summary, the question asks for the range of the function f(x) = (1-(1/x))^x, which is defined for x > 1. By using l'Hopital's rule, we can find that the limit of the function as x tends to infinity is 1/e. Therefore, the range of the function is 0 < x < 1/e. We can also use implicit differentiation to show that the derivative of the function is positive for all x > 1, which proves that the function is monotonic decreasing over this interval. Finally, we can use proof by contradiction to show that the function is not monotonic increasing over [1, ∞), which further supports our conclusion that the range is bounded
  • #1
Jimboza
1
0

Homework Statement


Calculate the range fo the function f:[1,infinite) ->R x maps to (1-(1/x))^x



Homework Equations


lny=xln(1-(1/x))
lhopitals rule.
Basic differentiation of logs



The Attempt at a Solution


I have worked on this for many many hours now and I am going nuts. I have found the limit as x tends to infinite to be 1/e by a few applications of l'hopitals rule. And at 1 f(x)=0
So if we know that the function never goes above 1/e before it tends to the limit the range is 0<x<1/e (should be less or equal sign at front but i don't know how to do that)

So i differentiated once using implicit differentiation. In tis most simplified form
dy/dx=y((1/(x-1))-ln(1-(1/x))) So if i can prove the gradient is always positive ,which i know it is :'( , for x>1 which is all we care about here we know the line NEVER goes above the limit before decreasing to approach the limit. But i cannot prove this. I have used inequalities to no avail and unfortunately graphing won't prove that 1/(x-1) +ln(1-1/x) is always greater than 0 as the question implies only using calculations. Please assist me in proving the range is indeed boudn between the limit and 0 for x>1 I would appreciate it so damn much.
 
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  • #2
When x >0 the function is monotonic decreasing. When x<0 the function is also monotonic decreasing. What happens at x=0? what happens when x goes to - infinity?
 
  • #3
Jimboza said:
In tis most simplified form
dy/dx=y((1/(x-1))-ln(1-(1/x)))
That is of the right form but it is not correct.

So if i can prove the gradient is always positive ,which i know it is :'( , for x>1 which is all we care about here we know the line NEVER goes above the limit before decreasing to approach the limit.
When you get the correct derivative, you should be able to ascertain a few of facts about the derivative. What happens to the derivative as x tends to 1 or to infinity? Is the derivative finite over (1+ε,∞) for any positive ε? What does the latter say about the continuity of your f(x)?

Prove it by contradiction. Assume the function is not monotonic increasing over [1,∞). This means that (and you should prove this) the second derivative must have zero in (1,∞). Does it?
 
  • #4
This is an ASSIGNMENT question from the MATH1901 course at the University of Sydney, due in on the 25th of this month. It violates PF guidelines to not even inform the Helper's that this is an assignment question that counts towards your final mark.
 
  • #5
Gib Z said:
This is an ASSIGNMENT question from the MATH1901 course at the University of Sydney, due in on the 25th of this month. It violates PF guidelines to not even inform the Helper's that this is an assignment question that counts towards your final mark.

Hi, could you clarify where this is communicated? I looked at the "Rules" page linked from every page, and found the following:

Homework Help:
<snip things irrelevant to this discussion>

We do not support cheating in any form: Do not ask for solution manuals, answers to exams, or instructor's manuals. Every school and instructor has their own policies or honor codes on what constitutes cheating, and it is up to the individual student to adhere to those policies when seeking help here. If you are in doubt as to whether you are permitted to seek help, consider erring on the side of caution and not asking for help.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

This seems to indicate that the student bears the responsibility for ensuring compliance with the ethics rules of his/her school. However, your post indicates that PF has its own guidelines which must be complied with. Where can these be found?
 

1. What is the formula for finding the range of (1-(1/x))^x from x=1 to infinite?

The formula for finding the range of (1-(1/x))^x from x=1 to infinite is y = 1/e to 1.

2. How do I find the range of (1-(1/x))^x from x=1 to infinite using a calculator?

To find the range of (1-(1/x))^x from x=1 to infinite using a calculator, you can graph the function and use the trace or table feature to find the y-values for different values of x. Alternatively, you can use a graphing calculator to find the y-values for a specific x-value by plugging in the value for x and solving for y.

3. What does the range of (1-(1/x))^x from x=1 to infinite represent?

The range of (1-(1/x))^x from x=1 to infinite represents the set of all possible output values for the function as x approaches infinity. It shows the limit of the function as x gets larger and larger.

4. How does the graph of (1-(1/x))^x from x=1 to infinite change as x increases?

As x increases, the graph of (1-(1/x))^x from x=1 to infinite approaches the horizontal line y=1. This means that the y-values of the function get closer and closer to 1, but will never actually reach 1.

5. Can the range of (1-(1/x))^x from x=1 to infinite be negative?

No, the range of (1-(1/x))^x from x=1 to infinite cannot be negative. As x approaches infinity, the function approaches the value of 1, which is positive. Therefore, the range of this function will always be between 1/e and 1, and cannot be negative.

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