Simple limit at infinity

1. Jan 23, 2007

Rasine

Find the limit as x-> -infinity for (x+(x^2+12x)^1/2)

so first of all..i multiply and divide by the conjugent then i get...

-12x/(x-(x^2+12x)^1/2)

i divide by x in both the nummerator and denominator to get ....
-12/1-(1+12/x)^1/2

so the 12/x goes to 0 and the squroot of 1 is 1 so it appears to be

-12/1-1 which is undefined....that is not right

where am i going wrong

2. Jan 23, 2007

arildno

Of course it is right. The function diverges.

3. Jan 23, 2007

Rasine

so the limit does not exist?
or it is undefined

how can i express that

4. Jan 23, 2007

arildno

It "equals" positive infinity, if you like.

5. Jan 23, 2007

mathman

Simplify:
(x^2+12x)^1/2 ~ x for large x, therefore your expression is ~ 2x

6. Jan 23, 2007

Gib Z

For large x, a polynomials value is determined by its leading term. Since x is large, x^2 + 12x is x^2. That to ^1/2 is just x. x+x, 2x. Substitution, it doesn't exist.

7. Jan 23, 2007

StatusX

The limit is as x goes to negative infinity. (x^2+12x)^1/2 looks like -x+const for x large and negative, so the series goes as x-x+const, and the limit is finite.

8. Jan 23, 2007

Gib Z

Sorry I didnt see the negative sign....Ill rethink that

9. Jan 24, 2007

dextercioby

The limit is -6.

Daniel.

10. Jan 24, 2007

HallsofIvy

Staff Emeritus
Dextercioby is, as usual, correct.

Replace the limit at $-\infty$ with a limit at $\infty$ by replacing x with -x:
$$lim_{x\rightarrow -\infty} x+ (x^2+ 12x)^{\frac{1}{2}}= lim{x\rightarrow\infty} (x^2- 12x)^{\frac{1}{2}}- x$$
Multiply "numerator and denominator" by the conjugate:
$$lim_{x\rightarrow\infty}\frac{x^2-12x-x^2}{(x^2-12x)^{\frac{1}{2}}+ x}= lim_{x\rightarrow\infty}\frac{-12x}{(x^2-12x)^{\frac{1}{2}}+ x}$$
Divide both numerator and denominator by x:
$$lim_{x\rightarrow\infty}\frac{-12}{(1-\frac{12}{x})^{\frac{1}{2}}+ 1}$$

Now it is obvious that the numerator is -12 and the denominator goes to 2.