# How to show that $0 \leq x \leq \frac{\epsilon}{1+\epsilon} \implies 1\leq \left(1+\frac{x}{n} \right)^{n}\leq \epsilon$

Show that

$$\forall n\geq 1,\quad \forall \epsilon > 0,\forall x \in \mathbb{R},$$

$$0 \leq x \leq \frac{\epsilon}{1+\epsilon} \implies 1\leq \left(1+\frac{x}{n} \right)^{n}\leq \epsilon$$

This question is related to that one Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $1\le \bigl(1+\frac{\alpha}{n}\bigr)^{n}\le \frac{1}{1-\alpha}$

My thoughts:

Since $$0 \leq x \leq \dfrac{\epsilon}{1+\epsilon}< 1$$ (because of $$\epsilon < \epsilon+1 \quad \forall \epsilon > 1$$) then by that question Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $1\le \bigl(1+\frac{\alpha}{n}\bigr)^{n}\le \frac{1}{1-\alpha}$ we've :

$$1\le \left(1+\dfrac{x}{n}\right)^{n}\le \dfrac{1}{1-x}$$

Still we've to prove that :

$$\dfrac{1}{1-x}\le 1+\epsilon$$

Or we have $$0 \leq x \leq \dfrac{\epsilon}{1+\epsilon} \implies 1-\dfrac{\epsilon}{1+\epsilon} \leq 1-x \leq 1 \implies \dfrac{1}{1+\epsilon} \leq 1-x \leq 1$$ which prove that last inequality

• Is my reasoning correct or is there another ways to prove that ?

Yes, it's correct, assuming the RHS is $1+\varepsilon$. Another way would be to observe that $\left(1+\frac xn\right)^n$ increases when you increase $x$, so you only have to prove it for $x=\varepsilon/(1+\varepsilon):$ $$\left(1+\frac{\varepsilon/(1+\varepsilon)}n\right)\le\frac1{1-\varepsilon/(1+\varepsilon)}=\frac{1+\varepsilon}{1+\varepsilon-\varepsilon}=1+\varepsilon$$