Question

solve for the horizontal asymptote:

lim (√((x^2)+7x-2)) +x
x->-∞

Answer 1

what tools do you have at your disposal?

Answer 2

a scientific calculator?

Answer 3

lol, i meant what kind of math can you use
i think the best bet is to multiply top and bottom by the conjugate of the numerator

Answer 4

its like university calculus? Lol
i tried conjugating by multiplying (√((x^2)+7x-2)) -x top and bottom

Answer 5

the conjugate is
\[\sqrt{x^2+7x-2}-x\]
when you multiply top and bottom you will get
\[\frac{x^2+7x+2-x^2}{\sqrt{x^2+7x-2}-x}\]
\[=\frac{7x+2}{\sqrt{x^2+7x+2}-x}\]

Answer 6

yup

Answer 7

now you want to compute the limit as \(x\to -\infty\)

Answer 8

i tried using the "divide everything by the highest power" method, and i got \[\frac{ 7+\frac{ 2 }{ x } }{ \sqrt{x ^{2}+7x+2}-x }\]

Answer 9

we can do this with our eyeballs, or do some calculation
eyeball method is to say that the numerator is degree 1, and is the denominator because \(\sqrt{x^2} -x\) looks like \(2x\) (since we are going to \(-\infty\) both are positive, then numerator is \(-7x\) so you get \(-\frac{7}{2}\)

Answer 10

ok we can do it that way too, but you have to divide the denominator by \(x\) as well

Answer 11

so inside the square root, i divide everything by X^2, and outside the square root, i divide the -x by x

Answer 12

yes but don't forget that you are going to minus infinity
so both the radical and(-x\) are positive, but the numerator will be negative
that is why it is \(-\frac{7}{2}\)

Answer 13

is there such thing as a -0?

Answer 14

nope

Answer 15

so 2/(-∞) is still 0

Answer 16

but i dont know what happens at the bottom

is it 1-0=1?

Answer 17

why do i divide the (-x) in the denominator by (-x)?

Answer 18

you don't

Answer 19

actually, if this is confusing, maybe it is best to divide everything by \(-x\)

Answer 20

ok let me try that

Answer 21

you have to keep track of the fact that it is the limit as x goes to minus infinity, not plus infinity
so both the radical in the denominator, and the term \(-x\) are positive, while the \(7x\) in the numerator is negative

Answer 22

it makes sense by dividing with (-x) but why?

Answer 23

because of what i wrote above

Answer 24

i am trying to think of a better way to explain it, but it is alluding me
that is why i like the eyeball method better
numerator is negative, denominator is positive,
numerator looks like \(-7x\) denominator looks like \(2x\)

Answer 25

but why is numerator negative? how do i determine that in other questions?

Answer 26

or perhaps a better way to explain would be this
take the negative of this whole thing, and then take the limit as x goes to \(+\infty\)

Answer 27

numerator is negative because you are going to minus infinity
x goes through numbers like -100,-1000,-1000000000 etc

Answer 28

so every time my limit is approaching x from the negative, i automatically assume the numerator to be negative?

Answer 29

well not every time, no
if it was say \(x^2\) then it would be positive no matter what

Answer 30

but in this case it is \(7x\) so it is negative for sure as \(x\to -\infty\)

Answer 31

oh yea!! that for sure. But when it is odd exponent, it will automatically be negative?

Answer 32

yes

Answer 33

THANK YOU SO MUCHHH!! :) that makes a whole lot sense now!

Answer 34

yw