Finding the Limit of x-> ∞ (5x^2-1)/(x^2)

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Problem statement
Find lim x-> infinity

(5x^2-1)/(x^2)

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Attempt at a solution
Just wanted to make sure I did this right.

Since the degree of the numerator is equal to the denominator does that mean that the limit is just the numerical coefficient of the leading term in the numerator and denominator so in this case it would be 5?
 
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Seat of the pants says so, but the proper way to show the limit is to split up the rational expression, do the necessary cancellations, and then evaluate the limits of the resulting expressions.
 
SteamKing said:
Seat of the pants says so, but the proper way to show the limit is to split up the rational expression, do the necessary cancellations, and then evaluate the limits of the resulting expressions.
In this case if you spilt it up you would get 5(infinity)-1/infinity ?
 
grace77 said:
In this case if you spilt it up you would get 5(infinity)-1/infinity ?

No, never. You want the limit of ##f(x) = (5 x^2 - 1)/x^2##. This can be written as
f(x) = \frac{5 x^2}{x^2} - \frac{1}{x^2}
Do you see now what happens?
 
Ray Vickson said:
No, never. You want the limit of ##f(x) = (5 x^2 - 1)/x^2##. This can be written as

f(x) = \frac{5 x^2}{x^2} - \frac{1}{x^2}

Do you see now what happens?
Yes I see it now! It would be equal to 5-0=5!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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