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How do I find the horizontal asymptote of (2)/(x^2-2x-3)?
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for large values of x, this is approximately: 2/x^2
so 2/(10^2)? which is .02
keep going; LARGE values of x 2/(1000000000000000000000000000^2)
Im gonna need a better claculator one sec
so y=0
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lol consider this logic: spose a pie is cut into 2 pieces, and you get 2 pieces of pie: 2/2 = 1 thats a large chunk of th epie right? now assume its cut into 10 pieces, 100 pieces, 124912304758580235 pieces, and each time you only get 2 slices are you getting more of the pie? or less of the pie each time?
yes, so y=0 :)
ok great! thank amistre64!
youre welcome
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