Evaluating limits when x -->infinity?

Simply look at the first term in both the nominator and deliminator. This simplifies to 1/x. The limit as x goes to infinity of 1/x is 0.

divide each term by the highest power of x in the given question.

lim x-> infinity (x^2/x^3 + 6x/x^3 + 8/x^3)/(x^3/x^3 + 2x/x^3 +1/x^3)

lim x-> infinity (1/x + 6/x^2 +8/x^3)/(1 + 2/x^2 + 1/x^3)

we evaluate the limits for the individual terms

lim x-> infinity (1/x + 6/x^2 +8/x^3)/(1 + 2/x^2 + 1/x^3) = (0 + 0+0)/(1 + 0 + 0) = 0/1 = 0

If you divide the top and bottom by x^3 (i.e. the largest power of x u can find in ur equation), then u get: [1/x^2 + 6/x^2 + 8/x^3]/[1 + 2/x^2 + 1/x^3]. Whatever power of x u have (power given by n), 1/x^n -> 0 as x->infinity. The value 1/x^n gets smaller and smaller the bigger x gets so it approaches 0. Your equation therefore approaches (1x0+6x0+8x0)/(1+2x0+1x0)=0/1=0. This works because you have a 1 on the bottom of the fraction so you don't get 0/0 which is undefined. There are other techniques to get around 0/0 if you can't get a non zero number on the bottom but they're not needed here obviously.

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