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How do you find #lim (5x+6)/(x^2-4)# as #x->oo#?

Answer 1

Christopher Agresta

The answer is #=0^-# and #0^+#

#lim_(x->-oo)(5x+6)/(x^2-4)=lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))#

#lim_(x->-oo)6/x=0#

#lim_(x->-oo)4/x^2=0#

#lim_(x->-oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->-oo)5/x=0^-#

#lim_(x->+oo)(5x+6)/(x^2-4)=lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))#

#lim_(x->+oo)(x(5+6/x))/(x^2(1-4/x^2))=lim_(x->+oo)5/x=0^+#

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Answer 2

Isabella Ackerman

To find the limit of (5x+6)/(x^2-4) as x approaches infinity, we can use the concept of limits. By dividing both the numerator and denominator by x^2, we can simplify the expression to (5/x + 6/x^2)/(1 - 4/x^2). As x approaches infinity, the terms with 1/x and 4/x^2 become negligible, resulting in the limit of (0 + 0)/(1 - 0). Simplifying further, we get 0/1, which equals 0. Therefore, the limit of (5x+6)/(x^2-4) as x approaches infinity is 0.

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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.

How do you find #lim (5x+6)/(x^2-4)# as #x-&gt;oo#?

How do you find #lim (5x+6)/(x^2-4)# as #x->oo#?