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

In summary: The limit is -6.In summary, the limit as x approaches negative infinity for (x+(x^2+12x)^1/2) is -6. This can also be expressed as the limit approaching positive infinity of (x^2 - 12x)^1/2 - x, which simplifies to -12/(1-12/x)^1/2 + 1 and ultimately approaches -6.
  • #1
Rasine
208
0
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
 
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  • #2
Of course it is right. The function diverges.
 
  • #3
so the limit does not exist?
or it is undefined

how can i express that
 
  • #4
It "equals" positive infinity, if you like.
 
  • #5
Simplify:
(x^2+12x)^1/2 ~ x for large x, therefore your expression is ~ 2x
 
  • #6
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
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
Sorry I didnt see the negative sign...Ill rethink that
 
  • #10
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.
 

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

1. What is a simple limit at infinity?

A simple limit at infinity is a mathematical concept that describes the behavior of a function as the input (x) approaches infinity. It is used to determine the ultimate value or behavior of a function as the input value becomes very large.

2. How is a simple limit at infinity different from a regular limit?

A simple limit at infinity is different from a regular limit because the input value (x) is approaching infinity, rather than a specific number. This means that the function is being evaluated at extremely large values, rather than specific values.

3. How do you find a simple limit at infinity?

To find a simple limit at infinity, you can use the rules of limits to simplify the function and then substitute infinity for the input value (x). If the simplified function approaches a finite value, then that is the simple limit at infinity. If the simplified function approaches infinity or negative infinity, then the simple limit at infinity does not exist.

4. Why is it important to understand simple limits at infinity?

Simple limits at infinity are important because they help us understand the overall behavior of a function as the input value gets larger and larger. This information can be useful in many applications, such as predicting long-term trends or behaviors in various systems.

5. Can a function have multiple simple limits at infinity?

Yes, a function can have multiple simple limits at infinity. This may occur if the function has different behaviors at different ranges of large input values. In this case, the simple limit at infinity would be different for each range of input values.

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