# What is the limit of a^x when a tend to infinity and x tend to 0?

What is the limit of [a]^{x} when [a]\rightarrow[/infinity] and[x]\rightarrow[/zero].It seem to me that it is divergent as lna,but I can not demontrate.It appears in the renormalization of Quantum Field Theory.
Thank you very much in advanced.

Landau
I am not sure how this limit should be interpreted formally.

We know that $\lim_{x\to\infty}x^{1/x}=1$, but it's not the same limit of course.

I like Serena
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According to http://en.wikipedia.org/wiki/Multivariable_calculus: [Broken]
"Since taking different paths toward the same point yields different values for the limit, the limit does not exist."

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The limit is one.
Lim(x,a -> 0,inf) a^x= L
lim x(lna)= ln(L)
lim (lna)/(1/x) = ln(L)
By l'hopital's, lim (1/a) / (-1/x^2) = ln(L)
lim(-x^2/a)=0=ln(L)
L=e^0=1

I like Serena
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If we set a = e^(1/x) and take the limit from x to zero from the positive side, we get:

$$\lim_{a,x \to \infty,0} a^x = \lim_{x \downarrow 0} (e^{\frac 1 x})^x = \lim_{x \downarrow 0} e^1 = e$$

In other words, it matters which path you take, so the limit does not exist.

Calculus textbooks say it this way: " $$\infty^0$$ is an indeterminate form ".

gb7nash
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Another way to look at the OP's problem is the rate at which the exponent approaches infinity and the base approaches 0. Depending on the rates, this could approach some finite number, +/-infinity or not exist.

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