Simple limit at infinity

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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
 

Answers and Replies

arildno
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Of course it is right. The function diverges.
 
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so the limit does not exist?
or it is undefined

how can i express that
 
arildno
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It "equals" positive infinity, if you like.
 
mathman
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Simplify:
(x^2+12x)^1/2 ~ x for large x, therefore your expression is ~ 2x
 
Gib Z
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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
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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
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Sorry I didnt see the negative sign....Ill rethink that
 
dextercioby
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The limit is -6.

Daniel.
 
HallsofIvy
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Dextercioby is, as usual, correct.

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

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

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