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How do you determine the limit of #(x² - 3x - 2)/(x² - 5)# as x approaches infinity?

Answer 1

#lim_(xrarrinfty)(x^2-3x-2)/(x^2-5)=1#

Given,

#lim_(xrarrinfty)(x^2-3x-2)/(x^2-5)#

Divide every term by the term with the highest degree in the denominator.

#=lim_(xrarrinfty)(x^2/color(blue)(x^2)-(3x)/color(blue)(x^2)-2/color(blue)(x^2))/(color(blue)(x^2)/color(blue)(x^2)-5/color(blue)(x^2))#

#=lim_(xrarrinfty)(1-3/x-2/x^2)/(1-5/x^2)#

Substitute #x=infty#.

#=lim_(xrarrinfty)(1-3/infty-2/infty^2)/(1-5/infty^2)#

Any constant divided by #infty# has a limit of #0# since the quotient would approach #0#.

#=lim_(xrarrinfty)(1-0-0)/(1-0)#

#=lim_(xrarrinfty)1/1#

#=1#

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Answer 2

Luke Alvarez

To determine the limit of (x² - 3x - 2)/(x² - 5) as x approaches infinity, we can divide the numerator and denominator by the highest power of x. This gives us (1 - 3/x - 2/x²)/(1 - 5/x²). As x approaches infinity, the terms with 3/x and 2/x² become negligible, and the limit simplifies to 1/1, which is equal to 1. Therefore, the limit of (x² - 3x - 2)/(x² - 5) as x approaches infinity is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.