Limit of [(7x^2)-x+11]/(4-x) as x->-infinity

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Homework Help Overview

The discussion revolves around finding the limit of the expression \(\frac{(7x^2)-x+11}{(4-x)}\) as \(x\) approaches negative infinity. The subject area is calculus, specifically limits and indeterminate forms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply L'Hopital's rule after identifying an indeterminate form but questions its applicability. Other participants suggest alternative methods, such as dividing by \(x\) to simplify the limit evaluation.

Discussion Status

Participants are exploring different methods to approach the limit problem. Some guidance has been offered regarding the use of L'Hopital's rule and the suggestion to simplify the expression by dividing both the numerator and denominator by \(x\). There is an acknowledgment of confusion surrounding limits involving infinity.

Contextual Notes

There is mention of a potential misunderstanding regarding the application of L'Hopital's rule and the nature of the limit as \(x\) approaches negative infinity. Participants are also reflecting on the implications of the quadratic term in the numerator.

kemmy
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Homework Statement



Find the Limit (as x-> -infinity) of [(7x^2)-x+11]/(4-x)

So I found that the limit is infinity/infinity indeterminate form so I tried to use L'hopital's to solve it.
So I took the derivative of (7x^2)-x+11 and got 14x-1
then for the derivative of 4-x I got -1

So the Limit as x->-infinity of 14x-1/-1 would be -infinity/-1 . So would the limit be infinity?

Apparently this answer is wrong. ...Is there a reason L'Hopital's wouldn't work here? How can I solve this?
Any help available would be much appreciated, thanks!
 
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The answer should be infinity, even without l'Hospital's, simply by virtue of the quadratic in the numerator. Check to you have the right equation or right way of checking the answer.
 
Why use L'Hopital? Divide both numerator and denominator by x. All fractions with x in the denominator will go to 0. What is left?
 
Okay... that does make sense. Thanks. I get so confused with Limits working with infinity I just always manage to get myself turned around. Thanks for all the help!
 
HallsofIvy said:
Why use L'Hopital? Divide both numerator and denominator by x. All fractions with x in the denominator will go to 0. What is left?

Denominator should be -1 right ? :smile:
unless you plug in infinity after you applied L'Hopital and divide both numerator and denominator by x.
 

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