Verified answer

The problem is with the signs. Epsilon and delta are positive.

Lim(x->10) (3-4x/5)=-5 if and only if for any E>0 there is d>0,

so that when |x-10|<d then |(3-4x/5)-(-5)|<E

Now you know that |3-4x/5-(-5)|<E. Must be true.

So |8-4x/5|<E....now factor out(4/5)

(4/5)|10-x|<E.... But |x-10| is the same as |10-x|

(4/5)|x-10|<E

|x-10|<(5/4)E

Now choose d= (5/4E)

|x-10|<d

Hoping this helps!