Math Is Fun Forum

the one in #140 doesn't exist because when we aproach it from two different paths it has different values.it is similair to the situation with one variable limits when the left and the right limit do not match up.

i said that the second limit you did exists.that is the one you did in post #143.

the first limit does not exist and the second one is correct but i tried to make a very bad joke.

hi ganesh

hi MIF

this is the first time since my registration that i have seen you online!

hi ganesh

hi Thanh Van

like i said that is the most i have figured out.i don't know what to do from that point on.

hi zee-f

14 and 16 are correct but 15 isn't.how did you get your answer and did you check to see if it is correct?

hi bob

about the perfect squares method:
let us take your example 9x^2+30x+25=
9(x^2+10x/3+25/9)=
9(x^2+10x/3+(5/3)^2)=
9(x+5/3)^2=
3^2(x+5/3)^2=
(3x+5)^2

also i did in school something ike this:
x^2-5x+6=
x^2-5x+(5/2)^2-(5/2)^2+6=
(x-5/2)^2-25/4+4=
(x-5/2)^2-1/4=
(x-5/2)^2-(1/2)^2
then we use the difference of squares:
(x-5/2-1/2)(x-5/2+1/2)=
(x-6/2)(x-4/2)=
(x-3)(x-2)

hi ganesh

hi ganesh

hi Thanh Van

i forgot that the solution can also be the constant function f(x)=C.

i started doing it by first assuming that the function is multiplicative.than i substituted y=-x.i got:
f(-x^2)+f(0)=f(x-x^2)+f(-x)
than by using the assumption that the function is multiplicative i got:
f(-1)f(x^2)+f(-1)f(0)=f(-1)f(x^2-x)+f(-1)f(x)
f(x)f(x)+f(x)f(0)=f(x)f(x-1)+f(x)
With the assumption that f(x)<>0:
f(x)+f(0)=f(x-1)+1
for x=1 we get:
f(1)+f(0)=f(0)+1
f(1)=1

that is the furthest i got.sry