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Given the definition

$$e = \lim_{h\rightarrow 0} (1+h)^{1/h}$$

I noticed that the limit

$$\lim_{h\rightarrow 0} (1+hx)^{1/h}$$

could be evaluated via substituting $u = hx$ in the following way.

$$\lim_{h\rightarrow 0} (1+hx)^{1/h} = \lim_{u\rightarrow 0} (1+u)^{x/u}=\left( \lim_{u\rightarrow 0} (1+u)^{1/u}\right)^{x} = e^{x}$$

I'm having a little trouble finding a reference stating under what conditions a substitution of this nature is valid. It seems to work out correctly here, but I don't know with confidence whether or not similar substitutions would work for other problems with guaranteed accuracy.

Therefore, I am wondering: under what circumstances does the following substitution generalization hold?

$$\lim_{x\rightarrow a} f(g(x)) = \lim_{u\rightarrow \underset{x \rightarrow a}{\lim} g(x)} f(u).$$

Trevor Kafka
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  • May be helpful https://math.stackexchange.com/questions/4545377/bringing-limits-inside-functions-when-limits-go-to-infinity/4545387#4545387 – zkutch Oct 15 '22 at 02:22

1 Answers1

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This holds if $f$ and $g$ are continuous at $g(a)$ and $a$ respectively. Continuity of $g$ implies $\lim_{x\to a}g(x) = g(a)$, and then continuity of $f$ and $f\circ g$ ensures $$\lim_{u\to g(a)}f(u) = f(g(a)) = \lim_{x\to a}f(g(x))$$

E G
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    That's definitely true, but limited in scope, as it can't even apply to the example shown here, since we don't have continuity at $h=0$. I think a more precise set of conditions are in need here. – Trevor Kafka Oct 15 '22 at 03:10