Proving limit of quadratic using formal definition without explicitly given f(x)

Question

How would you prove the following statement?:

Assume that $\lim_{x\to1}{f(x)=6}$. Prove using the epsilon-delta definition that $\lim_{x\to1}[5f(x)^2+ 3f(x)] = 198$.

I set up the following:
We have $\forall \epsilon_1$ in the reals, $\exists \delta_1$ such that $0<|x-1|<_1 ⇒ |f(x)-6|<\epsilon_1$
We want a $\delta$ such that $|5f(x)^2 + 3f(x) - 198|<\epsilon$
This becomes $|5f(x)+33| |f(x)−6| <\epsilon$, then, we have $|f(x)−6| <\frac{\epsilon}{|5f(x)+33|}$
Now, I am not sure how to continue and how to implement the $\delta = \operatorname{min}\{..., ...\} $thing

Answer 1

HINT:

If $|f(x)-6|<\varepsilon_1$ then $|f(x)|\le |f(x)-6|+6\le 6+\varepsilon_1$.

And, in particular, if you assume that $\varepsilon_1<1$, then $$|5f(x)+33|\le 5|f(x)|+33\le 5(6+\varepsilon_1)+33 < 68.$$

Answer 2

Note that\begin{align}5f^2(x)+3f(x)-198&=5f^2(x)-180+3f(x)-18\\&=5\bigl(f^2(x)-36\bigr)+3\bigl(f(x)-6\bigr)\\&=5\bigl(f(x)+6\bigr)\bigl(f(x)-6\bigr)+3\bigl(f(x)-6\bigr)\\&=\bigl(5f(x)+33\bigr)\bigl(f(x)-6\bigr).\end{align}So, take $\delta_0>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<1.$$Then, if $|x-1|<\delta_0$, you have $5<f(x)<7$. But\begin{align}5<f(x)<7&\iff25<5f(x)<35\\&\iff58<5f(x)+33<68\\&\implies\bigl|5f(x)+33\bigr|<68.\end{align}Now, given $\varepsilon>0$, take $\delta_1>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<\frac\varepsilon{68}$$and then$$|x-1|<\min\{\delta_0,\delta_1\}\implies\bigl|5f^2(x)+3f(x)-108\bigr|<\varepsilon.$$