Answer
See the proof below.
Work Step by Step
We have $$ |4x^2+2x+5-5| = |2x(x+1)| = 2 |2x+1| |x|=2 |2x+1| |x-0|$$
Since $|4x^2+2x+5 -5| $ is a multiple of $|x-0|$, then $|4x^2+2x+5 -5| $ is arbitrarily small whenever $ x $ is sufficiently close to $0$. Hence, we get $$\lim_{x\rightarrow 0}4x^2+2x+5=5.$$
