Question

Using the $\in -\delta$ definition prove that $\underset{x\to -2}{lim}(3x+8)=2$

Answer 1

The function $f\left(x\right)=3x+8$ is a polynomial and as such it is continuous every $x\in R$. Then $\underset{x\to 2}{lim}f\left(x\right)=f(-2)=3\times 2+8=-6+8=2$

To prove it by using the definition of limit,

$|f\left(x\right)-2|=|3x+8-2|=|3x+6|=3|x+2|$

For $x\in (2-f,2+f)$ with $f>0$, we have $|f\left(x\right)-2|=3|x+1|<3f$

Given any $e>0$, chose $f_e<e/3$ s.t

$x\in (-2-f_e,-2+f_e)\Rightarrow |f\left(x\right)-(-2)|<3f_e<\in$

which proves the result.