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Jul 26, 2020
Edited: Jul 26, 2020Limit[((1 + x)^(2/x) - e^2 (1 - Log (1 + x)))/x, x -> 0]
Is the answer e^2 or 0, me and my boi have two results.
Limit[((1 + x)^(2/x) - e^2 (1 - Log (1 + x)))/x, x -> 0]
Is the answer e^2 or 0, me and my boi have two results.
6 answers2 replies
I can't remember how many times I walked away from this but I kept coming back. I went around in circles for a long time.
I finally plotted a graph of [(1+x)^(1/x) - e]/x and there it was staring me in the face : -e/2 on the y-axis. times 2e to get -e^2. Combine with the first one to get 0.
Your intuitive way, I think, is more elegant. But here is the good old fashioned brute force method. Thanks again. Ian
Oh! Thank ya for your work! I'll just put your old fashioned method into my pocket, if you don't mind XD. And sorry for the late reply, something's wrong with my VPN, took me a while to fix it. Anyway, look forward to discussing more math problems with ya!! ;-)
@Ian Fowler In 2, are you sure it is in indeterminate form? plug 0 into the expression you had in 2, (1-ln(1+0))/0=(1-ln(1))/0=1/0 therefore, that bit also diverges
You are, of course, correct!! I was asleep at the wheel when I replaced ln(1) with 1 instead of 0. How embarrassing. I'll have another go tonight.
As it stand now, do we agree that we have (e^2 - e^2)/0 so the whole thing is 0/0 indeterminant?
@Ian Fowler yup
I think we cannot plug 0 as x into the expression, cuz it's 'x->0'.
Anyway here's my way, will it work? I hate limit questions...
I think you have nailed it in your photo. Good work. Sorry for the mess up on my first attempt. Ian
Limit[((1 + x)^(2/x) - e^2 (1 - Log (1 + x)))/x, x -> 0]