Find the Limit as n Approaches Infinity

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SUMMARY

The limit as n approaches infinity for the sequence defined by xn = (n^2 + log n)/(2n^3 - 1)^(1/2) can be evaluated by dividing both the numerator and denominator by n^2. This results in the expression lim (n to infinity) xn = lim (n to infinity) (1 + log n/n^2) / sqrt(2 - 1/n^4). As n approaches infinity, the numerator approaches 1 and the denominator approaches sqrt(2), leading to a final limit of 1/sqrt(2).

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Mattofix
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Please help, i take out n^2 top and bottom so end up with 0 as demominator...

Find lim (n to infinity) xn

xn = (n^2 + log n)/(2n^3 - 1)^(1/2)


...?
 
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You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.
 
if i divide the bottom by n^2 i get (1/n + 1/n^4)
 
sorry... (1/n - 1/n^4)^(1/2)
 
damn, sorry again, i mean (2/n - 1/n^4)^(1/2)
 
which becomes 0 as n tends to infinity?
 
\lim_{n\to \infty} x_n = \lim_{n\to\infty} \left( \frac{ 1+ \frac{\log n}{n^2} }{ \sqrt{ \frac{2n^3-1}{n^4}}} \right)

Consider separately, what is the numerator tending towards? How about the denominator?
 
Last edited:
Rationalize the denominator

Try rationalizing the denominator first. Then things get much easier.
 

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