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

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The limit as x approaches infinity for the expression (5x^2 - 1)/(x^2) is evaluated by recognizing that the degrees of the numerator and denominator are equal. The limit can be simplified to 5 - 0, resulting in a final value of 5. Proper evaluation involves splitting the rational expression and canceling terms appropriately. This confirms that the leading coefficient of the highest degree terms dictates the limit. Therefore, the limit is conclusively 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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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