How do I find the limit of this exponential expression?

Question

Can someone tell me in detail how to solve the following limit? I do have the solution to it from WolframAlpha but I cannot understand it:

$$\lim_{x\to\infty} \log\Bigl(1+\frac{2}{x}\Bigr)^x$$

Answer 1

Notice that the inside of the $\log$ is:

$$\lim_{x\to\infty}\left(1+{2\over x}\right)^x$$

Which is the limit definition of $e^x$:

$$e^x = \lim_{n\to\infty} \left(1+{x\over n}\right)^n$$

Where $x=2$. Thus the inside of the logarithm approaches $e^2$. Then, you can take the $\log$ from there.

Answer 2

You want $\lim_{x\to\infty} \log\Bigl(1+\frac{2}{x}\Bigr)^x =\lim_{x\to\infty} x\log\Bigl(1+\frac{2}{x}\Bigr) $.

But $\log(1+z) =\int_0^z \dfrac{dt}{1+t} \le z $ since $1+t \ge 1$ and $\log(1+z) =\int_0^z \dfrac{dt}{1+t} \ge \dfrac{z}{1+z} = \dfrac{z+1-1}{1+z} = \dfrac{z+z^2-z^2}{1+z} = z-\dfrac{z^2}{1+z} $ since $1+t \le 1+z $.

Therefore $x\log\Bigl(1+\frac{a}{x}\Bigr) \le x\frac{a}{x} =a $ and $x\log\Bigl(1+\frac{a}{x}\Bigr) \ge x\left(\frac{a}{x}-\frac{(a/x)^2}{1+a/x}\right) =a-\frac{a^2}{x+a} $.

Therefore, since $\lim_{x \to \infty} \frac{a^2}{x+a} =0$, $\lim_{x\to\infty} x\log\Bigl(1+\frac{a}{x}\Bigr) =a$.

Answer 3

Beside the use of the classical formulae involving the exponential functions, just consider $$ A=\log \left(\left(1+\frac{a}{x}\right)^x\right)=x\log \left(1+\frac{a}{x}\right)$$ and remember that, using Taylor series (or even equivalents) $$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$$ Since $x\to\infty$, make $\epsilon=\frac a x$ to get $$A=x \left(\frac{a}{x}-\frac{a^2}{2 x^2}+O\left(\frac{1}{x^3}\right) \right)=a-\frac{a^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ whci shows the limit and also how it is approached.