What is the limit of #((5u^4)+8)/((u^2)-5)((7u^2)-1)# as x goes to infinity?

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Answer 1 · 2015-09-21T17:48:56

It is not clear what the intended question is.

I shall assume that the variables should not be #u# and #x#
(if they should be different, there is not enough information to determine the limit.)

#lim_(urarroo)((5u^4)+8)/((u^2)-5)((7u^2)-1) = lim_(urarroo)(((5u^4)+8)((7u^2)-1))/((u^2)-5)#

# = lim_(xrarroo)(35u^6-5u^4+56u^2-8)/(u^2-5) = oo#

(The degree of the numerator is greater than the degree of the denominator, so the limit is either #oo# or #-oo#.)

If the intended limit is #lim_(xrarroo) (5x^4+8)/((x^2-5)(7x^2-1))#

Then we have:

#lim_(xrarroo) (5x^4+8)/((x^2-5)(7x^2-1)) = lim_(xrarroo) (5x^4+8)/(7x^4-36x^2+5)#

# = (x^4(5+8/x^4))/(x^4(7-36/x^2+5/x^4)) = 5/7#