Question

lim as x approaches negative infinity of x+sqrt(x^2+2x)

Answer 1

So I multiplied by the conjugate and got \[\frac{ 2x }{ x-\sqrt{x^2+2x} }\] not sure how to proceed from there.

Answer 2

is that algebra right? let me check

Answer 3

I think so.

Answer 4

it maybe it is supposed to be -2x on top?

Answer 5

yeah, supposed to be -2x.

Answer 6

still unsure what to do from there though

Answer 7

no you are right

Answer 8

divide by \(x\) i think

Answer 9

oh yeah top should be \(-2x\) right
now i think you can do it if you divide by \(x\)

Answer 10

ok so then \[\frac{ -2 }{ 1-\sqrt{x^2+2x}/x }\]

Answer 11

now if i put infinity in i get 1 - infinity on bottom so it is -2 over -infinity so negatives cancel and left with 2 over infinity so limit is 0?

Answer 12

no hold on let me try this algebra, i think the denominator is going to be \(2\)

Answer 13

can u show me how please

Answer 14

yeah lets just look at the denominator

Answer 15

you have
\[\lim_{x\to -\infty}1+{\frac{\sqrt{x^2+2x}}{x}}\]

Answer 16

nope i wrote it wrong

Answer 17

\[\lim_{x\to -\infty}1-{\frac{\sqrt{x^2+2x}}{x}}\]

Answer 18

now here is the thing, when you bring the \(x\) inside the radical you have to change the sign, since x is going to minus infinity not plus infinity

Answer 19

but its squared

Answer 20

yeah, the numerator is positive and the denominator is negative (since x is going to minus infinity)

Answer 21

i mean of this portion
\[\frac{\sqrt{x^2+2x}}{x}\]

Answer 22

ok so how do you know which x is dominant term to plug infinity into

Answer 23

or negative infintyy

Answer 24

if you want to be very precise, you change the sign and bring the x in the denominator as \(x^2\) inside the radical and make it
\[\lim_{x\to \infty}\sqrt{1+\frac{2}{x}}\]

Answer 25

which is pretty clearly \(1\) making the whole denominator \(2\) and your final answer \(-1\)

Answer 26

oh ok that makes sense

Answer 27

hope so
good luck!