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.

Question

Limit[((1 + x)^(2/x) - e^2 (1 - Log (1 + x)))/x, x -> 0]&#010;&#010;Is the answer e^2 or 0, me and my boi have two results.

anbowang48 · Accepted Answer

@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

Ian Fowler · Answer

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

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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.