Calculus Section 2.6 [ATTACH=full]3212[/ATTACH]

lim(e^(1/x)) as x->0 from the left note: e^(1/x)= .....so, if x->0 we have so, left limit is zero but if x->0 from the right, limit is infinity

The sqrt{e} when the index is 0 = e^(1/0) when replacing x with 0. I thought 1/0 = undefined. No? The graph explains it slightly better but what you said concerning x approaching 0 is not too clear.

If I replace x with 0 for sqrt{e}, is this not e^(1/0)? This means e^(undefined). I understand the graph but not what you said about sqrt{e} as x tends to 0 from the left.

I know that, Mira. As x tends to 0 from the left or right, we replace x with 0, right? So, sqrt{e} when x is the index leads to e^(1/x). Yes? If x = 0, do we or do we not get e^(1/0)? If this is true, we get e^(undefined). No?

The sqrt{e} = e^(1/2). No? The denominator of the exponent represents the index of the square root. No?