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

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The limit of the expression (5x^2 - 1)/(x^2) as x approaches infinity is definitively 5. This conclusion arises from the fact that the degrees of the numerator and denominator are equal, allowing the limit to be determined by the ratio of the leading coefficients. The proper method involves rewriting the expression as (5x^2/x^2) - (1/x^2), which simplifies to 5 - 0, confirming that the limit is indeed 5.

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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!
 

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