What is the Process to Find the Limit at Infinity of (x+(x^2+12x)^1/2)?

[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

 

[arildno]

Of course it is right. The function diverges.

 

[Rasine]

so the limit does not exist?
or it is undefined

how can i express that

 

[arildno]

It "equals" positive infinity, if you like.

 

[mathman]

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

 

[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.

 

[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.

 

[Gib Z]

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

 

[dextercioby]

The limit is -6.

Daniel.

 

[HallsofIvy]

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.