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Supposed I have $\lim_{x \to a}f(x)=-\infty.$ and $\lim_{x \to a}g(x)=c.$

Prove that $\lim_{x \to a}(f(x)+g(x))=-\infty.$

By the epsilon delta definition I know that for every $M<0$, i have:

$f(x)<M$ whenever $0<\left|x-a\right|<\delta_1$

and for every $\epsilon>0$ , i have

$\left|g(x)-c\right|<\epsilon$ whenever $0<\left|x-a\right|<\delta_2$

In order to prove the result I need to choose a $\delta_3$ such that given $Q<0$,

$f(x)+g(x)<Q$ whenever $0<\left|x-a\right|<\delta_3$

May I ask how I go about affixing the expression for $M$ and $\epsilon$ such that I will end up with $Q$ that will be negative by nature? Some hints will be appreciated!

R R
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Soby
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2 Answers2

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As you stated you have the ability to choose $M$ and $\varepsilon$ however you want. Intuitively we know we can make $f$ as large and negative as we want, and we can make $g$ as close to $c$ as we want. In particular we have a way of bounding $g$. So put a bound on $g$ and then make $f$ smaller than $Q$ minus the bound.

R R
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Fix $Q$ where $Q<0$.

Since $\lim_{x\to a}f(x)=-\infty$ then $\exists \delta_1$ such that $f(x)<Q-(c+1)$ whenever $|x-a|<\delta_1$

Since $\lim_{x\to a}g(x)=c$ then $\exists \delta_2$ such that $|g(x)-c|<1$ whenever $|x-a|<\delta_2$, that is $g(x)<1+c$ whenever $|x-a|<\delta_2$.

Now, set $\delta=\min\{\delta_1,\delta_2\}$.

When $|x-a|<\delta$, $f(x)+g(x)<Q-(c+1)+(c+1)=Q$.

Thus, $\lim_{x\to a} f(x)+g(x)=-\infty.$

QED.

Hasan Saad
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