Calculate the range fo the function f:[1,infinite) ->R x maps to (1-(1/x))^x
Basic differentiation of logs
The Attempt at a Solution
I have worked on this for many many hours now and im 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 dont 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 wont 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.