If $f(x) \rightarrow 0$ as $x \rightarrow a+ $ and $g(x) \geq 1$, $x \in \mathbb R $, then $\dfrac{g(x)}{f(x)} \rightarrow \infty$, $x\rightarrow a+$

Question

This is true right?

My approach:

If $f(x) \rightarrow 0$ as $x \rightarrow a+ $ then $a < x < a+\delta \rightarrow |f(x)| < \epsilon$

$f(x) \leq |f(x)| < \epsilon \rightarrow \dfrac{1}{f(x)} > \dfrac{1}{\epsilon}$

Let $M = \dfrac{1}{\epsilon}$

Then,

$\dfrac{g(x)}{f(x)} \geq \dfrac{1}{f(x)} > \dfrac{1}{\epsilon} = M $

Thus,

$\dfrac{g(x)}{f(x)} > M $ whenever $a < x < a+\delta$

Answer 1

No, this only holds if $f(x)>0$. If the function approaches 0 but bounces between positive and negative, then your limit will not exist as it will oscillate between positive and negative larger and larger values.

Example: $f(x)=x\cdot {\sin (\frac 1 x)} $, $g(x)=2$, $a=0$