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Why does $\lim_{x\to0} \left(1+\frac1x\right)^x=1$?

Beware that I am NOT asking about $ \lim_{x\to\infty} (1+\frac{1}{x})^x $, which I know equals to $e$.

When you draw it in GeoGebra or WolframAlpha, it tells us that this is true.

But WHY?

Any help (or better, proof) is welcome.

Blue
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RandomGuy
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2 Answers2

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$$\lim_{x \to 0}\left({1 + \frac{1}{x}}\right)^x=\lim_{x \to 0}e^{x\ln\left({1 + \frac{1}{x}}\right)}=^{t=\frac{1}{x}}\lim_{t \to \infty}e^{\frac{\ln(1+t)}{t}}=1$$

$$\lim_{t \to \infty}\frac{\ln(1+t)}{t}=\lim_{t \to \infty} \frac{1}{t+1}=0$$

Mostafa
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First, let's notice that $\lim_{u \to +\infty}\frac{u}{2^u} = 0$ (which should be obvious). This implies the existence of $U(\varepsilon > 0) : \forall u > U(\varepsilon), \frac{u}{2^u} < \varepsilon$.

Second, let's see that $\lim_{x \to 0}(\frac{1}{x})^x = 1$: if it were $1 + 2\varepsilon$, we could take $x = 2^{-u}, u > U(\log_2(1+\varepsilon))$ and get $(\frac{1}{x})^x = 2^\frac{u}{2^u} < 2^{\log_2(1+\varepsilon)} = 1 + \varepsilon$.

Finally, let's deal with $1+\frac{1}{x}$: $(1+\frac{1}{x})^{x} = [(x+1)(\frac{1}{x})]^x = (x+1)^x\cdot(\frac{1}{x})^x$ and limits of both multipliers as $x \to 0$ are $1$s.

(All the logic above assumes $x>0$: $(1 + \frac{1}{x})^x$ is usually considered ill-defined for real $x \in [-1,0]$; notice that sequences like $(-2n)^\frac{-1}{2n+1}$ converge to $-1$ for pretty much the same reason.)

The underlying logic of solution, of course, is backwards: first you see there should be little difference between $(1 + \frac{1}{x})$ and $\frac{1}{x}$, then you try some "large" $\frac{1}{x}$, like $2^n$ and get $2^\frac{n}{2^n}$, then you prove that the latter converges to 1 and retrace steps in a "proper" proof.

Abstraction
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