can someone tell me what's going on in this picture? Both sides seem identical to me, they both end up aproacching 1^infinity
>>8183128
>graph paper
Kill yourself.
>>8183128
They are not the same function.
how the fuck do you get euler's number from any of that?
Other has exponent 1/(1-x) and other 1/(x-1)
They are not the same
>>8183410
https://www.wolframalpha.com/input/?i=lim+x-%3E1+(2-x)%5E(1%2F(1-x))
>>8183128
There are two things you seem to be missing. The first is that though [math]\lim_{x \to 1} x-1 = 0 = \lim_{x to 1} 1-x[/math], in the second limit, you're considering [math]2-x[/math] to the power of [math]\frac{1}{1-x}[/math] and [math]\frac{1}{x-1}[/math]. But [math]\lim_{x \to 1} \frac{1}{1-x}[/math] and [math]\lim_{x \to 1} \frac{1}{x-1}[/math] do not exist (because you're dividing by a function which goes to 0).
The second is understanding how limit compositions work in general. If [math]\lim_{x \to a} g(x) = L[/math] and [math]\lim_{x \to L} f(x) = M[/math], then [math]\lim_{x \to a} f(g(x)) = M[/math] only when f is continuous at L. If you thought of [math]f(x) = (1+x)^{1/x}[/math], and [math]g(x) = 1-x[/math], you wouldn't be able to evaluate this limit because f is not continuous at [math]\lim_{x \to 1} g(x) = 0[/math].
>>8183451
To give a slightly informal argument about why the limits turn out the way they do, recall
[math]e^x = \lim_{n \to \infty} (1 + \frac{x}{n})^n[/math].
We're going to look at both limits from one direction to evaluate the actually limit of the function.
[math]\lim_{x \to 1^+} (2-x)^{1/(x-1)} = \lim_{n \to \infty} (1 + \frac{-1}{n})^n = e^{-1}[/math].
[math]\lim_{x \to 1^-} (2-x)^{1/(x-1)} = \lim_{n \to \infty} (1 + \frac{1}{n})^{-n} = \lim_{n \to \infty} \frac{1}{(1 + \frac{1}{n})^n} = e^{-1}[/math].
The idea behind the second function is similar.
>>8183128
thanks guys
>>8183268
Graph paper rocks, the only other acceptable paper is unruled ink jet paper.
