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:

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

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

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