Home > Calculus > Infinite Limits and Vertical Asymptotes

How do you find the limit of # (8x^2)/(4x^2-3x-1)# as x approaches infinity?

Answer 1

Isabella Ackerman

Divide terms on numerator/denominator by #x^2#

#((8x^2)/x^2)/((4x^2)/x^2-(3x)/x^2-1/x^2)=8/(4-3/x-1/x^2)#

as #xtooo,3/x" and " 1/x^2to 0#

#rArrlim_(xtooo)(8x^3)/(4x^2-3x-1)=8/4=2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

To find the limit of (8x^2)/(4x^2-3x-1) as x approaches infinity, we can divide both the numerator and denominator by the highest power of x, which is x^2. This gives us (8x^2)/(4x^2-3x-1) = (8)/(4-(3/x)-(1/x^2)). As x approaches infinity, both (3/x) and (1/x^2) approach zero. Therefore, the limit simplifies to (8)/(4-0-0) = 2. Hence, the limit of (8x^2)/(4x^2-3x-1) as x approaches infinity is 2.

By signing up, you agree to our Terms of Service and Privacy Policy

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.