0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.