My goal is to prove that

$$f(a) = \lim_{w \to 0} (1+aw)^{\frac{1}{w}} = e^a$$

without being too rigorous (just rigorous enough to convince myself that it really is true). Is the following method alright; or does it have flawed logic?

$$\frac{d}{da}f(a) = \frac{d}{da} (\lim_{w \to 0} (1+aw)^{\frac{1}{w}}) = \lim_{w \to 0} (\frac{d}{da} (1+aw)^{\frac{1}{w}})$$ $$= \lim_{w \to 0} (1+aw)^{\frac{1}{w} - 1}$$ $$ = \lim_{w \to 0} \frac{(1+aw)^{\frac{1}{w}}}{(1+aw)}$$ $$ = \frac{\lim_{w \to 0}(1+aw)^{\frac{1}{w}}}{\lim_{w \to 0}(1+aw)}$$ $$ = \lim_{w \to 0} (1+aw)^{\frac{1}{w}} = f(a)$$

Hence $\frac{d}{da}f(a) = f(a)$. Apart from the zero function, the only function that satisfies this property is $e^x$. Can we then conclude that $f(a) = e^x$?

  • $\begingroup$ If your question is whether from $f'=f$ and $f(0)=1$ we can conclude that $f=\exp$, the answer is yes. $\endgroup$ Feb 28, 2017 at 7:57
  • $\begingroup$ If you feel confident about interchanging the order of the two limiting processes $\frac{d}{da}$ and $\displaystyle\lim_{w\to0}$ then it looks fine. Otherwise your second equality certainly requires justification. $\endgroup$ Feb 28, 2017 at 7:59
  • $\begingroup$ You can not interchange order of differentiation and taking limits without any analysis. Moreover your approach still requires to solve a differential equation so it looks more like an overkill. $\endgroup$
    – Paramanand Singh
    Feb 28, 2017 at 12:29

1 Answer 1


You can also prove it by substitution with $aw=t$ :
$$\lim_{t\to 0}(1+t)^\frac{a}{t}=\lim_{t\to0}\left((1+t)^\frac{1}{t}\right)^a=\left(\lim_{t\to0}(1+t)^\frac{1}{t}\right)^a=e^a$$


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