verify $\lim_{x\rightarrow0}(4x^2+2x+5)=5$ Verify:
$$\lim_{x\rightarrow0}(4x^2+2x+5)=5$$
On a simple linear function it's easy to use the limit definition "$|f(x)-L|$ becomes arbitrarily small" but it won't work in this situation. But I'm supposed to use that definition somehow. The function won't factor. I'm not sure what to do.
We haven't learned limit laws yet so my professor wouldn't accept that.
EDIT: Changed my limit definition. Got it mixed up.
 A: Given any $\epsilon > 0$, let $\delta = \min\{1, \epsilon/6\} > 0$. Then observe that if $0 < |x - 0| < \delta$, then:
\begin{align*}
|(4x^2 + 2x + 5) - (5)|
&= |4x^2 + 2x| \\
&= |x(4x + 2)| \\
&= |x||4x + 2| \\
&< \frac{\epsilon}{6}|4x + 2| &\text{since } |x - 0| < \delta \leq \frac{\epsilon}{6} \\
&\leq \frac{\epsilon}{6}(|4x| + |2|) &\text{by the triangle inequality} \\
&= \frac{\epsilon}{6}(4|x| + 2) \\
&< \frac{\epsilon}{6}(4(1) + 2) &\text{since } |x - 0| < \delta \leq 1 \\
&= \frac{\epsilon}{6}(6) \\
&= \epsilon
\end{align*}
as desired.
A: I think a slightly simpler choice than the other answers is $\delta = \min\{\sqrt{\epsilon/8}, \epsilon/4\}$. Then if $|x - 0| = |x| < \delta$,
$$
|(4x^2 + 2x + 5) - 5| = |4x^2 + 2x| \le |4x^2| + |2x| < \epsilon/2 + \epsilon/2 = \epsilon.
$$
A: Have you shown in class that polynomials are continuous everywhere? If not, see this post for a sketch of a proof by induction on the polynomial degree of this extremely useful fact.
Next, if a function $f$ is continuous at $c$, then 
$$\lim_{x\to c} f(x) = f(c).$$
This is exactly the definition of continuity. So you can just plug in the 0 into your polynomial.
