Prove limit by Cauchy definition $\lim_{x \to -4} \frac{x-10}{x+6} = -7$

Question

I need to prove: $ \lim_{x \to -4} \frac{x-10}{x+6} = -7 $ so I said let ${\epsilon}$ > 0. the instructor told us to find a $δ$ such that for any x that $0<|x-a|<δ$ the following happens: $|{f(x) - L| = |\frac{x-10}{x+6} + 7| <\epsilon }$ and I got to this:

$ {8* |\frac{x+4}{x+6}| < \epsilon } $

Now how do I choose a correct $δ$ value to say that this expression is correct for any ${\epsilon}$? We just finished sequences and this new definition is really confusing to me. Thanks

Answer 1

$f(x):=8\dfrac{|x-(-4)|}{|x+6|}$.

Consider

$|x-(-4)|<1$, then

$-1 < x-(-4)<1$; $-1-4<x< 1-4$;

$-5<x<-3$; finally $-5+6<x+6<-3+6$;

$f(x):=|\dfrac{x-10}{x+4}+7| =8\dfrac{|x+4|}{|x+6|} <$

$8\dfrac{|x-(-4)|}{1}$.

Choose $\delta = \min (1,\epsilon/8)$.

Then $|x-(-4)|<\delta$ implies

$f(x)<8|x-(-4)| <8\delta <\epsilon$.