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

Answer 1

Cameron Alexander

#1/3#

#"divide terms on numerator/denominator by "x^2#

#=(x^2/x^2+(2x)/x^2-1/x^2)/(3/x^2+(3x^2)/x^2)=(1+2/x-1/x^2)/(3/x^2+3)#

#rArrlim_(xtooo)((x^2+2x-1)/(3+3x^2))#

#=lim_(xtooo)((1+2/x-1/x^2)/(3/x^2+3))#

#=(1+0-0)/(0+3)=1/3#

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Answer 2

Cameron Alexander

To find the limit of (x^2+2x-1)/(3+3x^2) 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 (1/x^2 + 2/x - 1/x^2)/(3/x^2 + 3). Simplifying further, we get (1 + 2/x - 1/x^2)/(3/x^2 + 3). As x approaches infinity, the terms 2/x and 1/x^2 become negligible, since they approach zero. Therefore, the limit simplifies to 1/3.

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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.