Prove that $0<\lim\limits_{x\to \infty}f(x)<1\implies \lim\limits_{x\to \infty}\big(f(x)\big)^x=0$

Question

Prove that $$0<\lim_{x\to \infty}f(x)=l<1\implies \lim_{x\to \infty}\big(f(x)\big)^x=0.$$

---

I'm looking for a delta - epsilon proof for this.

Starting from the LHS, I got to $$\forall\varepsilon_1 >0,\; \exists m>0,\;\forall x>m,\; (l-\varepsilon_1)^x<\big(f(x)\big)^x<(l+\varepsilon_1)^x$$

But I can´t find a way to manipulate that inequality to arrive at $$\forall\varepsilon>0, \; \exists m'>0,\; \forall x>m',\; -\varepsilon<\big(f(x)\big)^x<\varepsilon$$ ($\varepsilon$ maybe as a function of $\varepsilon_1$).

---

Any hints or solutions are more than welcomed. Thanks in advance.

Answer 1

As a hint: since $0<\lim_{x\to\infty}f(x)<1$, you can especially find some $B<1$ and some $m>0$ so that $0<f(x)<B$ for all $x>m$.

Therefore, when $x>m$, we get $0<f(x)^x<B^x$. Can you see how to finish this off?

Answer 2

Hint: It might be easier to prove that $$ \lim_{x\rightarrow \infty } x \ln f(x) = -\infty$$

Note that, for large enough $x$, $f(x)$ is positive and less than $1$ (hence $\ln f(x)$ is negative).