Question

take lim x approaching -infinity
(1-2x)/(sqrt(x^2+x)) for example, why would we only divide the bottom by x^2 and not the top?

Answer 1

eyeball problem

Answer 2

think of it as
\[\frac{-2x}{x}\]

Answer 3

if you want to do it the math teacher way, divide numerator and denominator by \(x\) but that is mostly a waste of time

Answer 4

\[\frac{\frac{1}{x}-2}{\sqrt{1+\frac{1}{x}}}\]

Answer 5

confused

Answer 6

where?

Answer 7

the 1 in the numerator is not important

Answer 8

neither is the \(x\) in the square root in the denominator

Answer 9

why not

Answer 10

it is determined entirely by the highest degree in the top and bottom

Answer 11

just like
\[\lim_{x\to \infty}\frac{2x^2+3x-4}{3x^2+5x-20}=\frac{2}{3}\]

Answer 12

i get that , but take this other one for example , lim x approaching infinity (x^2-3x+7)/(x^3+10x-4), you would divide the top and bottom by x^3, but in the one i showed you first, the top is divided by x?

Answer 13

i thought the highest one in the denominator was the one used to divide

Answer 14

i would do none of those things
in the example you just wrote the degree of the denominator is larger than the degree of the numerator, making the horizontal asympote \(y=0\)
bigger denominator means smaller number (small as in close to zero)

Answer 15

if you want to divide by \(x^3\) you will get the same answer

Answer 16

if it like x^2/x^3 and x is approaching infinity , it would be 0

Answer 17

yes

Answer 18

ok, but in the first one, if you divide by x^2 on top and bottom, you wont get -2

Answer 19

OOOHO i see the problem

Answer 20

you have in the denomiantor \(\sqrt{x^2+x}\) not \(x^2+x\)

Answer 21

then when you divide by \(x\) it comes inside the radical as a \(x^2\)

Answer 22

i.e.
\[\frac{\sqrt{x^2+x}}{x}=\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}\]

Answer 23

so when you divide top and bottom you get literally
\[\frac{\frac{1}{x}-\frac{2x}{x}}{\sqrt{1+\frac{1}{x}}}\]

Answer 24

ok so this only works when there is a square root in either bottom or top, you can divide it by lets say the top x and the bottom x^2?

Answer 25

no

Answer 26

you divide top and bottom by \(x\)

Answer 27

but as is said, the x comes inside the radical as a \(x^2\)

Answer 28

\[\frac{\sqrt{x^2+x}}{x}=\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}\]

Answer 29

ok, one last thing, lets change the top x with x^3 , what would we divide it by

Answer 30

if the numerator was \(x^3\) then the limit would be infinite, since the degree of the top would be larger than the degree of the bottom

Answer 31

btw your answer is \(2\) since you are taking the limit as x goes to \(-\infty\) not \(\infty\)

Answer 32

oh okay thanks

Answer 33

yw