How to solve limits when x approaches minus value?

Question

How to solve the following limit?

I can't think of a way to find $\delta$ values because x approaches a minus value . $$\lim_{x\to -3} x^2+5x+6=0$$

Answer 1

We want to show that for every $\epsilon$ there is a $\delta$ so that if $0<|x+3|<\delta$, then $|x^2+5x+6|<\epsilon$. Suppose $|x+3|<\delta$, then $|x^2+5x+6| = |x+3||x+2|<\delta |x+2|$. If we additionally assume $\delta < 1$, then we have: $$|x+3|<1$$ $$-1 < x+3 < 1$$ $$-2 < x+2 < 0$$ $$|x+2|<2$$

So $|x^2+5x+6|<2\delta$. This means that if we choose $\delta = \min(1,\frac{\epsilon}{2})$, we have:

$$|x^2+5x+6| = |x+3||x+2|<\delta \cdot 2 \leq \epsilon$$

Answer 2

HINT : $$ x^2+5x+6=(x+2)(x+3). $$

Answer 3

I think you should rewrite the expression as $(x+3)(x+2) = (x-(-3))(x-(-3)-1)$ and then use the fact that $|x-(-3)|< \delta$