Question

I'm trying to understand limits at infinity better. This problem is giving me fits:
limit at negative infinity of (x + sqrt(x^2 + 2x + 3))

Answer 1

Here is a propperly drawn equation:
\[\lim_{x \rightarrow -\infty} (x + \sqrt{x^{2} + 2x + 3})\]

Answer 2

use componendo dividendo

Answer 3

I do not understand what you mean.

Answer 4

gimmick is to multiply by
\[\frac{x-\sqrt{x^2+2x+3}}{x-\sqrt{x^2+2x+3}}\]

Answer 5

or u can do 2 substitutions:
first, put x by -x and make x-> infinity
then put x = 1/y
to make y->0

Answer 6

i mean to say if you do it like this
\[x+\sqrt{x ^{2}+2x+3}=\frac{x+ \sqrt{(x ^{2}+2x+3) }+1 }{x+\sqrt{x ^{2}+2x+3}-1 }\]
take x common and then apply the limits

Answer 7

"x common" ??

Answer 8

Oh, and BTW at my current point in the course, we're not supposed to use L'Hopital's rule. There is a possibility my TA made a mistake putting this in the study guide if this requires L'Hopital's rule to solve.

Answer 9

it does not require L'Hospitals rule

Answer 10

I would start with satellite73's 'gimmick'

Answer 11

I did. And I get \[\lim_{x \rightarrow -\infty} \frac{ -2x - 3 }{ x - \sqrt{x^2 + 2x + 3} }\]

Answer 12

My problem comes because I do not understand what to do with \[\sqrt{-\infty^2 + 2(-\infty) + 3}\]

Answer 13

factor out an \(x^2\) from under the radical

Answer 14

by my suggested procedure you need to divide both numerator and denominator with x

Answer 15

@Zarkon would that then basically cause \[\sqrt{2(- \infty) + 3}\] to be ignorable?

Answer 16

my answer is 2

Answer 17

the answer is not 2

Answer 18

@harsh314 the correct answer is \[-\infty \] I just don't know how to get there.

Answer 19

are you sure it is \(-\infty\)

Answer 20

I get -1

Answer 21

Yes. That is what my TA provided as the answer.

Answer 22

sorry i did a small mistake

Answer 23

that is not correct

Answer 24

it is -1

Answer 25

wolfram agrees with me...
http://www.wolframalpha.com/input/?i=Limit [x%2Bsqrt%28x^2%2B2x%2B3%29%2Cx%2C-infinity]

Answer 26

Oh! You're right. http://www.wolframalpha.com/input/?i=limit+at+-+infinity+%28x+%2B+sqrt%28x%5E2+%2B+2x+%2B+3%29%29

Well crap, now what do I do?

Answer 27

type Limit[x+sqrt(x^2+2x+3),x,-infinity] into wolfram alpha

Answer 28

do what I said above...factor out an \(x^2\) from under the radical

Answer 29

but i got two answers one is 3 and the other is minus one due to the square root

Answer 30

using the usual topology you can only get one answer for a limit

Answer 31

Ah, @Zarkon I do get -1 now. You never answered my question above though-- the reason that works is because it causes the square root in the denominator to be ignorable?

Answer 32

it is not exacty compulsory to get one answer only plz try it by the one i suggested

Answer 33

\[\frac{ -2(- \infty) - 3 }{ -\infty - -\infty \sqrt{2(-\infty) - 3} }\]

Answer 34

That would simplify to:\[\frac{ 2\infty }{ -2\infty }\]

Answer 35

the reason why iam saying so is if you try to put a limit of x tending to 1 on the function\[\sqrt{x}\] then you get two answers one is -1 and the other is+1

Answer 36

\[\lim_{x \to-\infty} \frac{ -2x - 3 }{ x - \sqrt{x^2 + 2x + 3} }\]
\[=\lim_{x \to-\infty} \frac{ -2x - 3 }{ x - \sqrt{x^2}\sqrt{x + \frac{2}{x} + \frac{3}{x^2}} }\]
\[=\lim_{x \to-\infty} \frac{ -2x - 3 }{ x - |x|\sqrt{x + \frac{2}{x} + \frac{3}{x^2}} }\]
our x<0 (it is going to -infinity so |x|=-x

\[=\lim_{x \to-\infty} \frac{ -2x - 3 }{ x +x\sqrt{x + \frac{2}{x} + \frac{3}{x^2}} }\]

factor out the x in the bottom ...simplify...take limit

Answer 37

small tyop

Answer 38

typo ;)

Answer 39

My knowledge breaks down at \[\sqrt{-\infty}\] Could you please explain what happens to that?

Answer 40

\[=\lim_{x \to-\infty} \frac{ -2x - 3 }{ x+x\sqrt{1 + \frac{2}{x} + \frac{3}{x^2}} }\]

Answer 41

Ok, @Zarkon I see your correction. So, you would have the "square root of 1 - very small - very small"? Which would then be 1?

Answer 42

yes..\(\sqrt{1}=1\)

Answer 43

\[=\lim_{x \to-\infty} \frac{ -2x - 3 }{ x(1+\sqrt{1 + \frac{2}{x} + \frac{3}{x^2}} )}\]

multiply top and bottom by 1/x

\[=\lim_{x \to-\infty} \frac{ -2 - (3/x) }{ 1+\sqrt{1 + \frac{2}{x} + \frac{3}{x^2}} }\]

\[=\frac{-2+0}{1+1}=-1\]

Answer 44

OK, I think I understand now. Thank you so much!

Answer 45

no problem