Trouble finishing my solution for a limit

Question

Given $k>0$ prove that: $\lim_{x \to \infty} (1+ \frac{k}{x})^x=e^k$.

The ideas I have tried are writing the limit as follows:

$$\lim_{x \to \infty} \left(1+ \frac k x \right)^x = \lim_{x \to \infty} e^{x \log(1+ \frac{k}{x})}.$$

Later I tried to show that $\lim_{x \to \infty} x \log(1+ \frac{k}{x})=k$.

And I tried to use Bernoulli´s inequality to get something like this:

$$x \log \left(1+ \frac k x \right) \le x \log\left(\left(1+ \frac 1 x\right)^k\right) =xk \log\left(1+ \frac 1 x\right)<k.$$

I believe that last part is true since $\forall x>0 \rightarrow \log(1+\frac{1}{x}) < \frac{1}{x}$.

Answer 1

Well, you can try this,

$$\lim_{x\rightarrow\infty}x\log(1+\frac{k}{x})=\lim_{x\rightarrow\infty}\frac{\log(1+\frac{k}{x})}{\frac{1}{x}}$$ [$\frac{0}{0}$ form]

By L'Hôpital's rule,

$$=\lim_{x\rightarrow\infty}\frac{-\dfrac{k}{x^2}\dfrac{1}{1+\dfrac{k}{x}}}{-\dfrac{1}{x^2}}=k$$

Answer 2

It depends a bit on how you define the exponential and the logarithm. In any case, the limit still holds for any $k$, positive, zero, or negative.

One way would be to use that $\log(1+t)=t+o(t^2)$. Then $$ x\log\left(1+\frac kx\right)=k+o(x^{-1})\to k. $$ Then taking the exponential, and using that it is continuous, you get your limit.