Question

Find G if......
limx→−2 (x^2+(G−1)x+G)/(x2) =1

Answer 1

x+2 on the bottom not x2

Answer 2

G is a constant?
And that the difference of G and 1 in that one group of parenthesis right?

Answer 3

\[\lim_{x \rightarrow -2}\frac{x^2+(G-1)x+G}{x+2}=1\]

Answer 4

I think something maybe wrong with this question
because if this limit were to exist you should have 0/0
and if you have 0/0 you should also be able to use l'hosptal rule
and I'm getting two different G's

Answer 5

Thats the equations and i need to find G, but I'm not sure where to start

Answer 6

well you should have 0/0 when pluggin -2

Answer 7

start there

Answer 8

the denominator would be 0 so i cant solve for G

Answer 9

For \[\lim_{x \rightarrow -2}\frac{x^2+(G-1)x+G}{x+2} \text{ \to exist } \]
you must have 0/0 when pluggin in -2

Answer 10

so \[(-2)^2+(G-1)(-2)+G=0 \]

Answer 11

what about the bottom (x+2), what do i do with that?

Answer 12

does x+2 = 0 when x=-2?

Answer 13

Yes

Answer 14

And also l'hopstail will work
just differentiated wrong :p

Answer 15

then we dont need to do anything with it at the moment ...

Answer 16

ok...

Answer 17

\[\frac{(-2)^2+(G-1)(2)+G~=~0}{(-2)+2~=~0}\]
solve for G

Answer 18

ugh, dropped a -

Answer 19

\[L=\lim_{x \rightarrow a}\frac{f(x)}{g(x)}\]
If L is a number we do have to have
f(a)=0 and g(a)=0
or
f(a)=some constant and g(a)=some constant not equal to 0

Answer 20

obviously we have the first case since g(a) does equal 0
so we have to find when f(a)=0

Answer 21

Though f(a)=0 and g(a)=0 might give other conclusions than L being a number

Answer 22

but here we are assuming it will give us an L that exists since L=1

Answer 23

4 - 2G +2 + G = 0

6 - G = 0

Answer 24

Try @amistre64 's answer there
and you will see it will give you the limit is 1

Answer 25

but im confused of the -2+2=0 and what we do with it since its on the bottom causing an error because we cant divide by 0

Answer 26

it only causes an error of x+2 is NOT a factor of the toip

Answer 27

*top

Answer 28

We are basically trying to chose a G such that we have a hole at x=-2 and not a vertical asymptote

Answer 29

if (x+2) is a common factor, we can essentially cancel it out with algebra

Answer 30

to get a hole we have to have 0/0 and not c/0 where c is not equal to 0

Answer 31

\[\frac{(x+2)~f(x)}{(x+2)}\]
when x=-2, top and bottom go to 0

Answer 32

we want to write
\[\frac{x^2+(G-1)x+G}{x+2}=\frac{(x+r)(x+2)}{(x+2)}=x+r \]

Answer 33

oops beat me to it

Answer 34

ok, so G=-6?

Answer 35

yes

Answer 36

wait no

Answer 37

6

Answer 38

it was half right :)

Answer 39

oh i see where i went wrong, so G=6 and that is my answer?

Answer 40

you want to see another way

Answer 41

yes please, if you dont mind

Answer 42

\[\frac{x^2+(G-1)x+G}{x+2}=\frac{(x+r)(x+2)}{(x+2)}=x+r \]
So when is \[x^2+(G-1)x+G=(x+r)(x+2)\]
\[x^2+(G-1)x+G=x^2+(r+2)x+2r\]
so G-1=r+2 and G=2r
replace G with 2r in first equation
2r-1=r+2
2r-r=1+2
r=3
so G=2r=2(3)=6

Answer 43

ok thank you both very much!!

Answer 44

There is also another way if you know l'hospital

Answer 45

\[2x+(G-1)=1 \text{ as } x \text{ approaches } -2 \\2(-2)+(G-1)=1 \\-4+G-1=1 \\G-5=1 \\ G=6 \]

Answer 46

@amistre64 check out my pretty system of linear equations way :P

Answer 47

ok, i have not learned that yet but it helps to see the answer

Answer 48

i had concidered Lhopping first, but you had already started with the long way ;)

Answer 49

the long way is pretty too

Answer 50

tkhunny suggest using long division

Answer 51

you could have done that too

Answer 52

or synthetic division also

Answer 53

tkhunny is never to be trusted .... never ever ever!

Answer 54

which is basically the first way

Answer 55

@tkhunny i spose its fair if you make it a tag :)

Answer 56

x
-------------------------------
x+2| x^2+(G-1)x+G
-(x^2+2x)
---------------
(G-1-2)x+G


so then
x+(G-3)
-------------------------------
x+2 | x^2+(G-1)x+G
-(x^2+2x)
------------------
(G-3)x+G
-((G-3)x+2(G-3))
----------------
G-2(G-3)

That remainder is G-2(G-3)
=G-2G+6
=-G+6
And the remainder is suppose to be 0
so -G+6=0 when G=6

Answer 57

I may not have understood the question. Are you sure you typed it the same the second time? This is why we like @amistre64

Answer 58

im devious ... lol

Answer 59

hey, @freckles and @amistre64 where did you get the 'r' from??

Answer 60

\[x^2+bx+c=(x+r_1)(x+r_2)\]

Answer 61

where r1 and r2 are the zeros of x^2+bx+c

Answer 62

just like if we had a cubic
\[x^3+bx^2+cx+d=(x+r_1)(x+r_2)(x+r_3)\]

Answer 63

where r is a complex number since not all quadratics are factorable over the reals

Answer 64

we knew we wanted \[x^2+(G-1)x+G \text{ to factor as } (x+2)(x+r)\]
so we could cancel the x+2 on bottom
we just called it r to begin with because we didn't know what the other zero needed to be at the time

Answer 65

ok thanks!!