# Under which conditions can we algebraically manipulate around an $\approx$ sign?

I noticed the following, and have a more general question that stems from it.

The natural logarithm can be defined as follows. $$\ln x := \lim_{\epsilon \rightarrow 0} \frac{x^\epsilon - 1}{\epsilon}$$ If we consider the case where $$\epsilon \approx 0$$ (for some non-rigorous definition of $$\approx$$), then we have $$\ln x \approx\frac{x^\epsilon - 1}{\epsilon}$$ This can then be algebraically manipulated as if $$\approx$$ were $$=$$. $$\epsilon \ln x \approx x^\epsilon - 1$$ $$1 + \epsilon \ln x \approx x^\epsilon$$ $$x \approx \left( 1 + \epsilon \ln x \right)^{1/\epsilon}$$ $$e^{\ln x} \approx (1 + \epsilon \ln x)^{1/\epsilon}$$ $$e^u \approx (1 + \epsilon u)^{1/\epsilon}$$ Turning this back into the language of limits, we have $$e^u = \lim_{\epsilon \rightarrow 0}(1 + \epsilon u)^{1/\epsilon}$$ This is, of course, true, though the math that got there is not rigorous, and wouldn't be possible if we kept the limit in place.

Here's another similar observation.

$$e$$ can be approximated by a Taylor series. $$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$ If $$x \approx 0$$, we can write $$e^x \approx 1 + x.$$ This can be rearranged to $$e \approx (1+x)^{1/x},$$ which in the language of limits is $$e = \lim_{x \rightarrow 0} (1+x)^{1/x},$$ which is of course a true statement. This also works if we take more Taylor terms. $$e^x \approx 1 + x + \frac{x^2}{2} \Longleftrightarrow e = \lim_{x \rightarrow 0} \left( 1 + x + \frac{x^2}{2} \right)^{1/x}$$ $$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \Longleftrightarrow e = \lim_{x \rightarrow 0} \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \right)^{1/x}$$ $$\text{et cetera}$$ As far as I can tell all of these statements are, surprisingly, true.

I have a couple questions about what's going on here:

1. What exactly is this relation $$\approx$$ in the way that I am using here?
2. Under what circumstances are algebraic manipulations that normally are done with $$=$$ instead done with $$\approx$$ valid?
3. Perhaps all of this is just limits in disguise. If you began with $$\ln x = \lim_{\epsilon \rightarrow 0} \frac{x^\epsilon - 1}{\epsilon}$$, is there a way to algebraically transform it to $$e^x = \lim_{\epsilon \rightarrow 0}(1 + \epsilon x)^{1/\epsilon}$$ in a mathematically sound way?

I'm basically looking for more intuition about exactly what is happening and what circumstances are allowing these hand-wavy manipulations to generate true statements from other true statements.

• Limits do follow certain algebraic rules, but the transformation you seek is not one of them. In general one is better off doing calculus using rules of calculus rather than trying to convert it into some sort of fancy algebra. – Paramanand Singh Mar 8 '19 at 5:39
• Also note that $e=\lim_{x\to 0}(1+x+100000x^2)^{1/x}$ but we don't have $e^x\approx 1+x+100000x^2$. – Paramanand Singh Mar 8 '19 at 5:44
• @ParamanandSingh Yeah, I totally understand that these rules do not hold in general. I'm curious what conditions are necessary for usual algebraic rules to hold. – Trevor Kafka Mar 8 '19 at 16:41