Proving limit of quadratic using formal definition without explicitly given f(x) 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 $0&lt;|x-1|&lt;_1 ⇒ |f(x)-6|&lt;\epsilon_1$
We want a $\delta$ such that $|5f(x)^2 + 3f(x) - 198|&lt;\epsilon$
This becomes $|5f(x)+33| |f(x)−6| &lt;\epsilon$, then, we have $|f(x)−6| &lt;\frac{\epsilon}{|5f(x)+33|}$
Now, I am not sure how to continue and how to implement the $\delta = \operatorname{min}\{..., ...\} $thing
HINT:
If $|f(x)-6|&lt;\varepsilon_1$ then $|f(x)|\le |f(x)-6|+6\le 6+\varepsilon_1$.
And, in particular, if you assume that $\varepsilon_1&lt;1$, then $$|5f(x)+33|\le 5|f(x)|+33\le 5(6+\varepsilon_1)+33 &lt; 68.$$
Note that\begin{align}5f^2(x)+3f(x)-198&amp;=5f^2(x)-180+3f(x)-18\\&amp;=5\bigl(f^2(x)-36\bigr)+3\bigl(f(x)-6\bigr)\\&amp;=5\bigl(f(x)+6\bigr)\bigl(f(x)-6\bigr)+3\bigl(f(x)-6\bigr)\\&amp;=\bigl(5f(x)+33\bigr)\bigl(f(x)-6\bigr).\end{align}So, take $\delta_0&gt;0$ such that$$|x-1|&lt;\delta_0\implies\bigl|f(x)-6\bigr|&lt;1.$$Then, if $|x-1|&lt;\delta_0$, you have $5&lt;f(x)&lt;7$. But\begin{align}5&lt;f(x)&lt;7&amp;\iff25&lt;5f(x)&lt;35\\&amp;\iff58&lt;5f(x)+33&lt;68\\&amp;\implies\bigl|5f(x)+33\bigr|&lt;68.\end{align}Now, given $\varepsilon&gt;0$, take $\delta_1&gt;0$ such that$$|x-1|&lt;\delta_0\implies\bigl|f(x)-6\bigr|&lt;\frac\varepsilon{68}$$and then$$|x-1|&lt;\min\{\delta_0,\delta_1\}\implies\bigl|5f^2(x)+3f(x)-108\bigr|&lt;\varepsilon.$$