how do you solve the question in the picture?(Itwasntme)

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

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- Oct 6th 2009, 12:30 PMSneakyNew Limit questions
how do you solve the question in the picture?(Itwasntme)

**Moderator edit:**The question is $\displaystyle \lim_{x \rightarrow - \infty} (x + \sqrt{x^2 + 3x})$. - Oct 6th 2009, 12:46 PMtonio

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 - Oct 6th 2009, 12:59 PMSneaky
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 - Oct 6th 2009, 01:05 PMtonio
- Oct 6th 2009, 01:13 PMSneaky
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 - Oct 6th 2009, 01:18 PMSneaky
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? - Oct 6th 2009, 01:22 PMtonio