Showing $\lim_{x \to -3}|\frac{5-x}{3+x}| = +\infty$ using the definition of the limit

Question

Show: \begin{align}\lim_{x \to -3}\left|\frac{5-x}{3+x}\right| = +\infty\end{align}

My Answer:

For all $M > 0$ there is $\delta > 0$ , such that $|\frac{5-x}{3+x}| > M$ whenever $0 < x + 3 < \delta$. We've looking in inequalities for to can choose the $\delta$ appropriate: \begin{align} \left|\frac{5-x}{3+x}\right| > M \to \left((\frac{5-x}{3+x})^2\right)^{1/2} > M \to \left(\frac{5-x}{3+x}\right)^2 > M^2 \end{align}

But, I don't know how to continue.

Answer 1

Observe that $|5-x| = |8-(x+3)| \ge |8| - |x+3|=8 - |x+3| > 8 - a$. Thus $\left|\dfrac{5-x}{x+3}\right|= \dfrac{|5-x|}{|x+3|}> \dfrac{8-a}{a}> M\iff8-a> aM\iff 8 > a+aM = a(1+M)\iff a< \dfrac{8}{1+M}$. This tells you how to find the $\delta$ and you can take $\delta = \dfrac{8}{1+M}>0$,and your proof is done.

Answer 2

$ \forall T \in \mathbb{R} \exists \delta > 0: \forall 0< |x+3|< \delta: f(x) > T$