Question about limits at infinity with radicals.

In summary, the limit can be simplified by multiplying the numerator and denominator by 1/sqrt(x). This will result in the limit equaling 2/-2, which is equal to -1.
  • #1
7
0

Homework Statement


Find the following limit:
[tex]
\lim_{x \to \infty} \frac{2+\sqrt{(6x)}}{-2+\sqrt{(3x)}}
[/tex]

Homework Equations


n/a

The Attempt at a Solution


I know this shouldn't be that hard, but somehow I keep getting stuck on simplifying the equation. I think the first step is to multiply both the denominator and numerator by 1/x (which equals 1/sqrt(x^2). This would give me

[tex] \lim_{x \to \infty} \frac{\frac{2}{x}+\sqrt{\frac{6x}{x^2}}}{\frac{-2}{x}+\sqrt{\frac{3x}{x^2}}} [/tex]

And if I'm correct the squareroots should go to zero which leaves:
[tex] \lim_{x \to \infty} \frac{\frac{2}{x}}{\frac{-2}{x}} [/tex]
But wouldn't those on top of each other equal 0/0 too unless I can multiply by x for both so it equals 2/-2?

I am really just not sure of the correct way to go about simplifying this. Any help would be appreciated!
 
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  • #2
I would factor [itex]\sqrt{x}[/itex] out of both terms in the numerator and both in the denominator.

That would give you
[tex]\lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x}}\cdot \frac{2/\sqrt{x}+\sqrt{6}}{-2/\sqrt{x}+\sqrt{3}}[/tex]
 
  • #3
No, you can't do it this way. You can say the square roots "go to zero" and can be canceled out, but you are right to observe that by the same argument, the [tex]2/x[/tex] terms should "go to zero" as well.

Try multiplying the numerator and denominator by [tex]x^{-1/2}[/tex] rather than [tex]x^{-1}[/tex].
 
  • #4
Ok I understand multiplying the numerator and denominator by 1/sqrt(X) to obtain:

[tex]
\lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}}+{\sqrt{6}}}{\frac{-2}{\sqrt{x}}+\sqrt{3}}
[/tex]

From this point do I multiply by the conjugate of the numerator?

[tex]
\lim_{x \to \infty} \frac{(\frac{2}{\sqrt{x}})^2-{6}}{-(\frac{2}{\sqrt{x}})^2+\frac{2\sqrt{3}}{\sqrt{x}}+\frac{2\sqrt{6}}{\sqrt{x}}-\sqrt{18}}
[/tex]

This seems like quite the jumbled mess so maybe I'm wrong..
 
  • #5
Kstanley said:
Ok I understand multiplying the numerator and denominator by 1/sqrt(X) to obtain:

[tex]
\lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}}+{\sqrt{6}}}{\frac{-2}{\sqrt{x}}+\sqrt{3}}
[/tex]

From this point do I multiply by the conjugate of the numerator?
No, just take the limit. The first terms in the top and bottom go to zero.
Kstanley said:
[tex]
\lim_{x \to \infty} \frac{(\frac{2}{\sqrt{x}})^2-{6}}{-(\frac{2}{\sqrt{x}})^2+\frac{2\sqrt{3}}{\sqrt{x}}+\frac{2\sqrt{6}}{\sqrt{x}}-\sqrt{18}}
[/tex]

This seems like quite the jumbled mess so maybe I'm wrong..
 
  • #6
Ah duh I got so into thinking it had to be complicated I didn't even think to just find the limit right there. Thank you so much that was so much easier than I imagined!
 

1. What is the definition of a limit at infinity?

A limit at infinity is the behavior of a function as the input approaches positive or negative infinity. It represents the value that the function approaches as the input grows infinitely large or small.

2. How do you determine if a limit at infinity exists?

To determine if a limit at infinity exists, you can evaluate the limit as the input approaches infinity from both the positive and negative sides. If these values are equal, then the limit exists. If they are not equal, then the limit does not exist.

3. What is the limit at infinity of a radical function?

The limit at infinity of a radical function depends on the degree of the radical. For example, the limit at infinity of a square root function is 0, while the limit at infinity of a cube root function is either positive or negative infinity.

4. Can a limit at infinity with radicals be evaluated using algebraic methods?

Yes, a limit at infinity with radicals can be evaluated using algebraic methods such as multiplying by the conjugate or rationalizing the denominator. These methods can help simplify the expression and make it easier to determine the limit.

5. What is the difference between a limit at infinity and a limit at a finite number?

A limit at infinity represents the behavior of a function as the input grows infinitely large or small, while a limit at a finite number represents the behavior of a function as the input approaches a specific value. In other words, a limit at infinity looks at the long-term behavior of a function, while a limit at a finite number looks at the behavior near a specific point.

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