Prove directly from the definition that $\lim: \lim_{x\to -2} x^2 + x - 5 = -3$

Question

So I am a bit stuck on applying the limit definition for this problem. This is what I have so far:

$\lvert x - c \rvert < \delta $ then $\lvert f(x) - l \rvert < \epsilon$$ $$\lvert x + 2 \rvert < \delta$ $$\lvert x^2 + x -2 \rvert < \epsilon $$

which means $\lvert x -1 \rvert \lvert x + 2 \rvert < \epsilon $

$\lvert x + 2 \rvert < {\epsilon\over \lvert x -1 \rvert} $

let $\delta = 1 $ then: $\lvert x + 2 \rvert < \delta < 1 $ which means $-1 < x -1 < 1 $ $-4 < x -1 < -2 $ Thus the upper bound of $\lvert x -1 \rvert < -2 $ is $-2$

let $$\delta = \min(1, {\epsilon\over -2}) $$

But what is confusing me here is the fact that, how is $ \lvert x + 2 \rvert < {\epsilon\over -2} $

Because if $\epsilon$ is any number more than $0$, then ${\epsilon\over -2} $ will be negative and it can't be greater than $ \lvert x + 2 \rvert $.

Answer 1

Note that\begin{align}x^2+x-5-(-3)&=x^2-4+x+2\\&=(x-2)(x+2)+x+2\\&=(x-1)(x+2).\end{align}So, if $\lvert x+2\rvert<1$, then $\lvert x-1\rvert\leqslant\lvert x+2\rvert+3<4$. So, take $\delta=\min\left\{1,\frac\varepsilon4\right\}$. Then, if $\lvert x+2\rvert<\delta$, then$$\lvert x^2+x-5-(-3)\rvert<4\frac\varepsilon4=\varepsilon.$$