1. ## New Limit questions

how do you solve the question in the picture?

Moderator edit: The question is $\lim_{x \rightarrow - \infty} (x + \sqrt{x^2 + 3x})$.

2. Originally Posted by Sneaky
[IMG]file:///C:/DOCUME%7E1/HP_Owner/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]how do you solve the question in the picture?

First we multiply the expression by 1:

[x + Sqrt(x^2 + 3x)]*(x - Sqrt(x^2 + 3x)]/[x - Sqrt(x^2 + 3x)] =

[x^2 - x^2 - 3x]/[x - Sqrt(x^2 + 3x)] = -3x/[x - Sqrt(x^2 + 3x)]

Now we're going to multiply the above by 1 = (1/-x)/(1/-x)...why -1/x?
Because we're going to "put" 1/-x inside the square root in the denominator, but as x --> -oo we HAVE to put inside the root something positive! Said this, pay atention to the fact that 1/-x gets into the root
as 1/x^2 , so :

-3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

by arithmetic of limits and since 3/x --> 0 when x --> +/- oo.

You can now grab your personal calculator and input low values for x into the function, say x = -1,000 or x = -100,000, and convince yourself the limit is -3/2 indeed.

Tonio

3. i dont understand

-3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

that line at all

how do u get the -3/2

4. Originally Posted by Sneaky
i dont understand

-3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

that line at all

how do u get the -3/2

Hmmm...3/x --> 0 and thus 1 + 3/x --> 1 and thus Sqrt(1 + 3/x) --> Sqrt(1) = 1.

In conclusion 3/[-1 - Sqrt(1 + 3/x)] --> 3/(-1 - Sqrt(1)) = 3 /(-1-1) = 3/-2

Tonio

5. i got to the step

-3x (1/-x)
-------------------------
(x- sqrt(x^2 + 3x))(1/-x)

then for the numerator i simplify and get 3
but for the bottum part i dont see how you get -2
and why does 3/x = 0, why do you assume that?

why does

-sqrt(x^2 + 3x)
---------------- = -1 ?
-x

6. ok so
sqrt(x^2 + 3x)
----------------
x
with this i understand how to get one, so now i get

3
-------
-1 +/- 1

so its either -3/2 or 3/0

so how do you know its -3/2 not the other one?

7. Originally Posted by Sneaky
i got to the step

-3x (1/-x)
-------------------------
(x- sqrt(x^2 + 3x))(1/-x)

then for the numerator i simplify and get 3
but for the bottum part i dont see how you get -2
and why does 3/x = 0, why do you assume that?

Oh, dear god: who ever wrote such a nonsense?! I said 3/x --> 0, not equality! Meaning, of course, that x--> oo or x --> -oo, it's the same in both cases.

Now the denominator:

(x- sqrt(x^2 + 3x))(1/-x) = x*(-1/x) - (-1/x)Sqrt(x^2 + 3x) =

-1 - Sqrt([x^2 + 3x]/x^2) = -1 - Sqrt(1 + 3/x)

Rememebr that if a is a non-negative real number, the a*Sqrt(h) = Sqrt(ah) , and this is why we needed -1/x = 1/-x and NOT 1/x, because since
we're taking the limit as x --> -oo we're going to work with NEGATIVE x's, so in order to get somethinbg into a square root we took -1/x, which is POSITIVE (as 1/x is negative!).

Tonio

Pd. Read carefully this and my prior answers, work on them with pencil and paper, THINK...and after a good while of honest effort, if you don't succeed then write back.

why does

-sqrt(x^2 + 3x)
---------------- = -1 ?
-x
---------------------------------------------