How do you find the limit #(t+5-2/t-1/t^3)/(3t+12-1/t^2)# as #x->oo#?

1 Answer
Steve M
Nov 4, 2016

# lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = 1/3 #

Explanation:

I assume that you mean #lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) # and that the expression should not contain the variable #t#!!

Now, As #x->oo# then #1/x->0#

So, it would be better if we could replace #x# with #1/x#

# lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = lim_(xrarroo)(1/x(x+5-2/x-1/x^3))/(1/x(3x+12-1/x^2)) #

# :. lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = lim_(xrarroo)(1+5/x-2/x^2-1/x^4)/(3+12/x-1/x^3) #

# :. lim_(xrarroo)(x+5-2/x-1/x^3)/(3x+12-1/x^2) = (1+0-0-0)/(3+0-0)=1/3 #

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