I've been trying my hand at a 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 $f(x)=\frac{5x}{x-5}$. Using a formal proof, prove that $\lim\limits_{x \to 5^+}f(x)=\infty$.
From my own prior understanding, the definition of a right handed 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 $a<x<a+\delta$, 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 $a<x<a+\delta$.
Then we have $5<x<5+\delta$ or $0<x-5<\delta$.
And so, $\frac{5}{x-5}>\frac{5}{\delta}$.
Here's where I get stuck. If this was a typical reciprocal function (which I have dealt with), I could simply see that $\frac{5}{\delta}$ must equal $M$, and so $\delta=\frac{5}{M}$. However, I cannot seem to understand how I should manipulate my inequality so that the left side becomes $f(x)$, without having the right side have an $x$ term. Is there some step that I haven't yet been introduced to which I need to use or am I going about this proof the wrong way?
Thank you so much in advance for any advice!
Take $M>0$. You are after some $\delta>0$ such that$$5<x<5+\delta\implies\left|\frac x{x-5}\right|>M\left(\iff\frac x{x-5}>M\right).$$But (if $M>1$)\begin{align}\frac x{x-5}>M&\iff \frac 5{x-5}>M-1\\&\iff x-5<\frac 5{M-1}.\end{align}So, take $\delta=\dfrac 5{M-1}$. And if $M\leqslant1$, any $\delta$ will do.