Prove: Assume $f(x)\le g(x)$, if $\lim_{x\rightarrow x_0} f(x) =\infty$ then $\lim_{x\rightarrow x_0} g(x) =\infty$

Let $f,g$ be two functions defined in a punctured neighbourhood of $x_0$, and assume $f(x) \le g(x)$ If $\lim_{x\rightarrow x_0}f(x)=\infty$ then $\lim_{x\rightarrow x_0}g(x)=\infty$

First: what is this theorem called? Second: I'd appreciate some help proving it for some review.

Attempt:

$\forall M \in R$ $\exists \delta \gt 0$

$$0 \lt |x-x_0| \lt \delta_1 \Rightarrow f(x)\gt M$$

and we have to show :

$$0 \lt |x-x_0| \lt \delta \Rightarrow g(x)\gt M$$

But my issue is, do we assume that the limit of g(x) at $x_0$ is $\infty$? Or do we assume it's $L \in R$ until we prove otherwise?

I'm going to try and say that g(x) has a limit "L"

$\forall \epsilon \gt$ $\exists \delta_{0,1,2} \gt 0$

$$0\lt |x-x_0|\lt \delta_1 \Rightarrow |g(x)-L|\lt\epsilon$$

$$|g(x)-L|\lt\epsilon \Rightarrow -\epsilon+L\lt g(x)\lt L+\epsilon$$

Is it right to take $\delta=min(\delta_1, \delta_2)$? Or is that not necessary?

And since $f(x) \le g(x)$ and $f(x)\gt M$:

$$M\lt f(x) \le g(x) \lt \epsilon +L$$

This is where I'm the most uncertain (not that Im certain of any earlier steps). Is it enough to show that $M \lt g(x)$? or do I need to now pick an epsilon based off of M to finish off the proof?

• Consider this, since $f(x)\leq g(x)$ so $\lim_{x\to x_0} f(x)\leq \lim_{x\to x_0} g(x)$. What can you conclude knowing about the limit of $f(x)$? Jun 14 '16 at 16:52
• Notice that you didn't cover all cases. You have only covered the case in which $g$ has a limit at $x_0$. There are more cases, such as $g$ is bounded but doesn't have a limit at $x_0$
– NSA
Jun 14 '16 at 16:54
• Also, notice that within $|x-x_0|<\delta$ you have that $M<f(x) \leq g(x)$, and there you have it.
– NSA
Jun 14 '16 at 16:56

But my issue is, do we assume that the limit of $g(x)$ at $x_0$ is $\infty$?

No, if you did that then you would assume what you are trying to show.

Now let $M>0$ be given, you then want to find a $\delta$ such that

$$0 \lt |x-x_0| \lt \delta \Rightarrow g(x)\gt M$$

Since you know that $\lim_{x\to x_0} f(x) = \infty$, then (by definition) you have that there exists a $\delta_1$ such that

$$0 \lt |x-x_0| \lt \delta_1 \Rightarrow f(x)\gt M$$

But since $f(x) \leq g(x)$ this same $\delta_1$ will work for $g$ as well. So

$$0 \lt |x-x_0| \lt \delta_1 \Rightarrow g(x) \geq f(x)\gt M$$

You don't need to assume anything about $g(x)$ besides what is stated in the problem. There is no need to consider special cases.

• Thank you for the corrections. I'm glad I was at least in the right direction. Jun 14 '16 at 17:01