I've been trying my hand at a another practice problem for my first-year differential calculus class, but I cannot seem to make it past a certain step. The problem is as follows:
Let $f(x)=\frac{-x}{(x+4)^2}$. Using a formal proof, prove that $\lim\limits_{x \to -4}f(x)=\infty$.
From my own prior understanding, the definition of a limit tending to infinity is as follows (roughly):
Given $\lim\limits_{x \to a}f(x)$, for every $M>0$, there exists a $\delta$ such that if $0<|x-a|<\delta$, then $f(x)>M$.
Here is my attempt at the problem (I am currently stuck on deriving the relationship between $\delta$ and $M$)
Let $M>0$ be given. Suppose $0<|x-a|<\delta$.
Then we have $0<|x+4|<\delta$.
And so, $\frac{1}{(x+4)^2}>\frac{1}{\delta^2}$
Taking a look at our initial inequality, we have $0<|x+4|$ and so $x>-4$ or $x<-4$. Using $x<-4$, we have $-x>4$.
By $\frac{1}{(x+4)^2}>\frac{1}{\delta^2}$, it should follow that $\frac{-x}{(x+4)^2}>\frac{4}{\delta^2}=M$ and so $\delta=\frac{2}{\sqrt{M}}$.
Using this value for $\delta$, we can thus formulate a proper proof. However, that's not my question. My question is, is my derivation for the relationship between $\delta$ and $M$ correct? Is something that I am forgetting to account for, because when I go and graph the appropriate inequalities and equation using graphing software, there are some values for $x$ such that if $|x+4|<delta$ then $f(x)<M$, which shouldn't occur.
Any help would be greatly appreciated!