Prove directly from the definition that $\lim: \lim_{x\to -2} x^2 + x - 5 = -3$ 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 &lt; \delta $ then $\lvert f(x) - l \rvert &lt; \epsilon$$ $$\lvert x + 2 \rvert &lt; \delta$ $$\lvert x^2 + x -2 \rvert &lt; \epsilon $$ 

which means $\lvert x -1 \rvert \lvert x + 2 \rvert &lt; \epsilon $ 

$\lvert x + 2 \rvert &lt; {\epsilon\over \lvert x -1 \rvert} $

let $\delta = 1 $ then: $\lvert x + 2 \rvert &lt; \delta &lt; 1 $ which means 
$-1 &lt; x -1 &lt; 1 $
$-4 &lt; x -1 &lt; -2 $
Thus the upper bound of $\lvert x -1 \rvert &lt; -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 &lt; {\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 $.
Note that\begin{align}x^2+x-5-(-3)&amp;=x^2-4+x+2\\&amp;=(x-2)(x+2)+x+2\\&amp;=(x-1)(x+2).\end{align}So, if $\lvert x+2\rvert&lt;1$, then $\lvert x-1\rvert\leqslant\lvert x+2\rvert+3&lt;4$. So, take $\delta=\min\left\{1,\frac\varepsilon4\right\}$. Then, if $\lvert x+2\rvert&lt;\delta$, then$$\lvert x^2+x-5-(-3)\rvert&lt;4\frac\varepsilon4=\varepsilon.$$
If $0&lt;ε&lt;2+$$\frac{1}4$$,then: 
|f(x)-l|&lt;ε$ $\Rightarrow$ $|x^2+x-5+3|&lt;ε$ $\Rightarrow$$-ε&lt;x^2+x-2&lt;ε$ $\Rightarrow$ $(0&lt;x^2+x-(2-ε)) \land (x^2+x-(2+ε)&lt;0)$ $\Rightarrow$ $(x&lt;\frac{-1-\sqrt{1+4(2-ε)}}2,\frac{-1+\sqrt{1+4(2-ε)}}2&lt;x ) \land (\frac{-1-\sqrt{1+4(2+ε)}}2&lt;x&lt;\frac{-1+\sqrt{1+4(2+ε)}}2)$

By combining the inequalities we acquire two intervals containig 1 and -2 respectively.We are interested in the interval that for every 0&lt;ε&lt;2+1/4 contains -2, and then you are done.You can fill in the details.
If $ε&gt;2+$$\frac{1}4$$,then:
-ε&lt;x^2+x-2 $ holds true for every ε, and the other inequality simplifies to an interval that contains both 1 and -2.So you do basically the same work with the first case but you only have one interval to work with. I hope you found this helpful. I thank Martin Sleziak for the help with the edit.