What is the value of c that will make f continuous at x=3?

Question

QUESTION- what is the value of c that will make f continuous at x=3? 

 

(1/x-2)-1/x-3 if x>3 

 

cx+1 if x< or = 3 

Answer 1

If   f(x)=1/(x-2)-1/(x-3), when x>3

     f(x)=cx+1, x≤3

lim_x→3⁺f(x)=lim_x→3⁺(1/(x-2)-1/(x-3))=∞

lim_x→3^-f(x)=lim_x→3^-(cx+1)=1+c

 

THEN there is no such finite c!

 

BUT if f(x)=1/(x-2)-1/x-3, for x>3, then 

lim_x→3⁺f(x)=lim_x→3⁺(1/(x-2)-1/x-3)=-7/3

and in order f(x) to be continous at x=3 c should be

c+1=-7/3, c=-10/3.

 

 

Answer 2

(1/3 - 2) - 1/3 - 3 = 1/3 - 2 -1/3 - 3 = -2 - 3 = -5

 

c*3 + 1 = -5

 

c*3 = -6

 

c = -6/3 = -2