Infinite limit at a point epsilon-delta proof

Question

Couldn't find much of info about these kinds of limits so im posting this here.

I have a problem that looks as follows:

$\lim_{x \to 3^+} \frac{x + 3}{x-3} = \infty$

The definition for this particular problem would be:

$\forall M > 0$ $\exists \delta$, such that when $0 < x-3 < \delta$ then $f(x) > M$

right?

I approached the problem as follows:

$\frac{x+3}{x-3} > M$ $\iff$ $x-3 < \frac{x+3}{M}$

This is the part im getting stuck at. The $x+3$ is confusing me, i don't think i can choose $\delta = \frac{x+3}{M} $ since i have the term with $x$ on the fraction, so how would i continue from here?

Answer 1

$\frac {x+3} {x-3} >M$ can be written as $x+3 >M(x-3)$ or $(x-3)+6>M(x-3)$ or $(M-1)(x-3) <6$ or $x-3 <\frac 6 {M-1}$. Hence $o <x-3 <\frac 6 {M-1}$ gives $\frac {x+3} {x-3} >M$.

Answer 2

$\forall M > 0,\exists \delta > 0, s.t. 0 < x-3 < \delta \implies \frac {x+3}{x-3} > M$

This sets us up for a right hand limit and one that goes to infinity.

Now we just need to find a definition for $\delta$ such that the statement above holds.

Nothing keeps us from insisting that $\delta \le 1$

And therefore $7<x+3<8$

$\frac {x+3}{x-3}>\frac {7}{\delta}$

Let $\delta = \max (\frac{7}{M}, 1)$