How do you find the limit of #(n^4+6n^2)/(n^4-2n+3)# as n approaches infinity?

1 Answer
Alexander
Jul 23, 2016

#lim_(n->∞) (n^4+6n^2)/(n^4-2n+3) = 1#

Explanation:

Since this is also known as an infinite limit, and both the numerator and denominator are polynomials, we can divide each term by the highest power, namely by #n^4#.

Therefore, we have

#lim_(n->∞) (n^4+6n^2)/(n^4-2n+3) = lim_(n->∞) (cancel((n^4/n^4))+(6n^2)/(n^4))/(cancel((n^4)/(n^4))-(2n)/(n^4)+(3)/(n^4))#

#= lim_(n->∞) (1+6/(n^2))/(1-2/(n^3)+3/(n^4))#

Also, since we know that #lim_(n->∞) 1/(n^m) = 0 iff m>0#, we then have

#lim_(n->∞) (1+6/(n^2))/(1-2/(n^3)+3/(n^4)) = 1/(1) = 1#

We can also graph our original function to see if this limit makes sense.

graph{(x^4+6x^2)/(x^4-2x+3) [-10, 10, -5, 5]}

In this case, we can see that as #n->∞#, our function tends to #1#.

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