prove a statement with implies and limits using epsilon.

Question

If a function is that $\lim_{x \to 4} f(x)=7, \implies\lim_{x \to 4}[(f(x))^{2}+f(x)+1]=57$. how do I use the $\epsilon$ and $\delta$ definition to prove this. I know if $\lim_{x \to 4}f(x)=7,$ then $|f(x)-7|<\epsilon$. The implies, statement is a quadratic equation. If $f(x)=a$, then $\lim_{x \to 4}[(f(x))^{2}+f(x)+1]=57$ is the same to $a^{2}+a+1$. And, then I can make it $a^{2}+a-56$. But what to do now? Can I have some assistance?

Answer 1

Since $\lim_{x \to 4} f(x)=7$, $\forall \epsilon_0 > 0$ there exists $\delta_0$ such that when $|x-4| < \delta_0, |f(x)-7| < \epsilon_0$.

Now $\forall \epsilon >0$, we pick $\epsilon_0 = \min (1,\frac{\epsilon_0}{16})$, therefore $(f(x))^2+f(x)+1 > f(x) > 0$ when $|x-4|<\delta_0$. Then we set $\delta= \delta_0$, and if $|x-4| < \delta=\delta_0, 6 < f(x) < 8$, we have $$\left||(f(x))^2+f(x)+1| - 57\right| = |(f(x))^2+f(x)+1 - 57|\\ = (f(x)+8)|f(x)-7| < 16 |f(x)-7| \le \epsilon.$$