indeterminate forms(1^infinity)?

Question

why limf(x)^g(x)=1^infinity is e^lim[f(x)-1]*g(x) ????????

Answer 1

hmmm... well...

lim f(x)^g(x)
= lim e^ln[f(x)^g(x)]
= e ^ lim ln[f(x)^g(x)]
= e ^ lim { g(x) * ln|f(x)| }


that should be the exact expression... then it turns out that ...

ln|f(x)| is being substituted by f(x) - 1 ... this is simply the approximate substitution whenever you truncate the taylor series for ln|B| for the first term...

ln|B| = 0 + 1*(B-1) - 1*(B-1)^2 + 1/2*(B-1)^3 - ...
(expanded about B = 1...)
thus
ln|B| is approximately B - 1.

e ^ lim { g(x) * ln|f(x)| } is approximately
e ^ lim { g(x) * [f(x) - 1] }


§

Answer 2

Suppose that as x→c, f(x)→1. Then [x→c]lim f(x)^g(x) = e^([x→c]lim (f(x)-1) g(x))

Proof: Since f(x) → 1, we can assume that f(x) is positive in some neighborhood of c, and thus that ln f(x) is defined in some neighborhood of c. Then it follows from the continuity of e^x that:

[x→c]lim f(x)^g(x)
= [x→c]lim e^(ln (f(x)) g(x))
= e^([x→c]lim ln (f(x)) g(x))

So it suffices to show that [x→c]lim ln (f(x)) g(x) = [x→c]lim (f(x)-1) g(x). To do this we define an auxiliary function h as follows:

h(x) = {(f(x) - 1)/ln (f(x)) if f(x)≠1, 1 if f(x)=1}

Note that if f(x)≠1, h(x)*ln (f(x)) = (f(x) - 1)/ln (f(x)) * ln (f(x)) = f(x) - 1, and that if f(x) = 1, h(x)*ln (f(x)) = h(x)*ln 1 = h(x)*0 = 0 = f(x) - 1, so for any value of x, h(x) * ln (f(x)) = f(x) - 1. Thus:

[x→c]lim (f(x)-1) g(x)
= [x→c]lim h(x) ln (f(x)) g(x)
= ([x→c]lim h(x)) * ([x→c]lim ln (f(x)) g(x)).

So if we can but establish that [x→c]lim h(x) = 1, then it will follow immediately that [x→c]lim (f(x)-1) g(x) = [x→c]lim ln (f(x)) g(x) and thus that [x→c]lim f(x)^g(x) = e^([x→c]lim (f(x)-1) g(x)).

Now, from L'hopital's rule it follows that:

[y→1]lim (y-1)/ln y
= [y→1]lim 1/(1/y)
= 1/(1/1)
= 1

Which means by definition that ∀ε>0, ∃δ>0 s.t. 0<|y-1|<δ ⇒ |(y-1)/ln y - 1|<ε. Also we are given that [x→c]lim f(x) = 1, which means that ∀ε>0, ∃D such that D is a punctured neighborhood of c and x∈D ⇒ |f(x) - 1|<ε (note that I am using the more general topological definition of limits here to account for the possibility that c might be ∞ or -∞ instead of some real number, or even that f might be a function whose domain is a topological space other than the real numbers. In short, I am trying to show that this limit always holds, regardless of the nature of f). So we can say the following:

Let ε>0. Choose a δ>0 s.t. 0<|y-1|<δ ⇒ |(y-1)/ln y - 1|<ε. Then choose a punctured neighborhood D of c such that x∈D ⇒ |f(x) - 1|<δ (we can do this since such a neighborhood exists for any real number greater than 0, and δ is a real number greater than 0). Now, suppose x∈D. Then |f(x) - 1|<δ. If |f(x) - 1| = 0, then f(x) = 1, and so h(x) = 1, thus |h(x) - 1| = 0 < ε. If |f(x) - 1| ≠ 0, then it follows that 0<|f(x)-1|<δ, so |(f(x)-1)/ln (f(x)) - 1|<ε. But in this case, (f(x) - 1)/ln (f(x)) = h(x), so we have |h(x) - 1|<ε. Thus in either case, x∈D ⇒ |h(x) - 1|<ε. And since we can find such a punctured neighborhood for any ε>0, it follows that ∀ε>0 ∃D s.t. D is a punctured neighborhood of c and |h(x)-1|<ε, so [x→c]lim h(x) = 1. Q.E.D.

This was probably way more formal than you asked for. If your teacher gives a proof at all, it will only be for the case where f is a function from R to R (the odds of a high school calculus teacher knowing anything about the definition of a limit on an arbitrary topological space are small), and may involve additional assumptions, such as c being a real number and possibly that f and g are differentiable functions (although as you can see, those additional assumptions are NOT required). Note also that nowhere in the proof did we use the fact that g(x) → ∞. We could just as well have said that g(x) approaches -∞, or some real number, or even that it does not have a limit; as long as [x→c]lim (f(x)-1)*g(x) exists, so too will [x→c]lim f(x)^g(x), and they will have the same value. It is also not too difficult to show that [x→c]lim f(x)^g(x) implies the existence of [x→c]lim (f(x)-1)*g(x) as well (assuming still that f(x)→1). So this theorem actually applies to 1^anything.