functions - Show that $frac{1}{1+k}=frac{frac{1}{k}}{1+frac{1}{k}}leq ln(1+frac{1}{k})leqfrac{1}{k}$

Prove the following:
$$\frac{1}{1+k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}\leq \ln(1+\frac{1}{k})\leq\frac{1}{k}$$

I know I can prove it with induction if the values were naturals. However, the "problem" for me is that they're real.

Answer

For all $x \in \mathbb{R}$,
$$e^x \geq 1 + x$$
Taking log on both sides we get,
$$\ln (1 + x) \leq x, \forall x > -1$$
Substituting $x = \frac{1}{k}, k \notin [0, -1]$, we get,
$$\displaystyle{\ln \left(1 + \frac{1}{k}\right) \leq \frac{1}{k}}$$
Substituting $x = \frac{-1}{k + 1}, k \notin [0, -1]$, we get,

$$\ln \left(\frac{k}{k + 1}\right) \leq \frac{-1}{k + 1}\Rightarrow \ln \left(1 + \frac{1}{k}\right) \geq \frac{1}{k + 1}$$