Limit of $\lim_{x\to 0^+}(1+\frac{1}{x})^x$

Question

Can someone give me advice how to solve this limit $\lim_{x\to 0^+}(1+\frac{1}{x})^x$ ?

Answer 1

Rewrite using logarithm and $e$:

$$\left(1+\frac{1}{x}\right)^x = e^{x\log(1+\frac{1}{x})}$$

Then, since $e^x$ is a continuous function, we have,

$$\lim\limits_{x\to0^+}e^{x\log(1+\frac{1}{x})}=e^{\lim\limits_{x\to0^+}x\log(1+\frac{1}{x})}$$

So we examine the limit,

$$\lim\limits_{x\to0^+}x\log\left(1+\frac{1}{x}\right)=\lim\limits_{y\to\infty}\frac{\log(1+y)}{y}$$

and now just apply l'hopitals,

$$\lim\limits_{y\to\infty}\frac{\log(1+y)}{y}=\lim\limits_{y\to\infty}\frac{1}{1+y}=0$$

So the final answer is $e^0=1$

Answer 2

Note that for $y=\frac1x\to \infty$

$$\left(1+\frac{1}{x}\right)^x = \left(1+y\right)^{\frac1y}=e^{\frac{\log(1+y)}y}\to e^0= 1$$

indeed by l'Hospital or by standard limits $y=e^z-1\to \infty$ with $z\to \infty$

$$\frac{\log(1+y)}y=\frac{z}{e^z-1}\to 0$$