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How do you evaluate the limit #(x^2-x-6)/(x+2)# as x approaches #oo#?

Answer 1

The limit of a sum is the sum of the limits, provided the individual limits exist. Similarly with product, quotient, etc.

Let's manipulate for the expression to a more convenient form #(x^2-x-6)/(x+2)=((x^2-x-6)/x)/((x+2)/x)=((x^2/x-x/x-6/x))/((x/x+2/x))=(x-1-6/x)/(1+2/x)#

Then #lim_(xrarroo) ((x^2-x-6)/(x+2)) = lim_(xrarroo) ((x-1-6/x)/(1+2/x)) = (oo -1 -0)/(1+0)=oo#

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Answer 2

Cameron Alexander

To evaluate the limit of (x^2-x-6)/(x+2) as x approaches infinity, we can use the concept of asymptotes. As x approaches infinity, the highest power term in the numerator and denominator dominates the expression. In this case, the highest power term is x^2 in the numerator and x in the denominator.

Dividing both the numerator and denominator by x, we get (x^2/x - x/x - 6/x)/(x/x + 2/x). Simplifying this expression, we have (x - 1 - 6/x)/(1 + 2/x).

As x approaches infinity, the terms with 1/x become negligible, and we are left with (x - 1)/(1).

Therefore, the limit of (x^2-x-6)/(x+2) as x approaches infinity is simply x - 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.