What is the algebraic approach to finding limits approaching infinity?

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I'm having a hard time learning from the textbook, I know I can do this if someone just outlines what my goal is here... and what I can interpret from that goal.

The solutions handbook just makes seemingly random algebraic changes to the limit function and then tells me what the answer is, and it's a bit confusing.

Homework Statement



Find the limit approaching infinity of:

[3sqrt(x^3)] / [sqrt(2x^3)]


Homework Equations





The Attempt at a Solution



I don't know where to begin. Don't solve it for me, explain what I am looking to get through the algebraic changes, and I will post an attempted solution.

Thanks.
 
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what would be the limit of \frac{2x^2}{x^2}
as it went to infinity
 
It helps to state the problem clearly. You want the limit as x approaches infinity. Your first step should be to simplify the algebraic expression you are working with.
 
Alright, after looking through more examples it looks like they want me to try to get to n/x (x approaching infinity) which becomes zero.

I also wrote the problem incorrectly, 3x should be added to the numerator. So will post back with an attempt shortly.
 
1MileCrash said:
Alright, after looking through more examples it looks like they want me to try to get to n/x (x approaching infinity) which becomes zero.

I also wrote the problem incorrectly, 3x should be added to the numerator. So will post back with an attempt shortly.

Let's get this straight, the question is to find

\lim_{x\to \infty}\frac{3x+3\sqrt{x^3}}{\sqrt{2x^3}}

?
 
Yes, but I completely understand my goal now. I'm just dividing every term by the highest x power in the denominator and it's working out eventually as n/x (0). My answer is 3/sqrt(2).
 
\sqrt{x^3}= x^{3/2}
so this is
\frac{3x+3x^{3/2}}{\sqrt{2}X^{3/2}}= \frac{3}{\sqrt{2}}\frac{x+ x^{3/2}}{x^{3/2}}

The highest power of x is 3/2 so divide both numerator and denominator by x^{3/2} to get
\frac{3}{\sqrt{2}}\frac{x^{-1/2}+ 1}{1}

That gives exactly what you say. Good work!
 
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