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If $x$ is fixed within $[0,1]$, the limit of $n^3x^n(1-x)$ as n tends to infinity is $0$.

But how do I show this? $n^3$ tends to inifinity, but the other term tends to zero, and surely I cannot simply assume that their product tends to $0$?

And if I reason that because $x^n$ dominates $n^3$ therefore the above function tends to zero, this neglects the possibility that the function may converge to somwhere between zero and infinity.

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2 Answers 2

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There are lots of ways to do this, but for fixed $x<1$, the only variable is $n$, and I suggest rewriting the problem as $\lim_{t\to\infty}t^3/c^t$ where $c=1/x>1$, by calling the variable $t$ instead of $n$. Then top and bottom go to infinity, and you can use L’Hospital three times to get your result. You do need to know that the derivative of $c^t$ is $\log(c)c^t$, and use the fact that the logarithm is positive.

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When $x = 0$ or $x = 1$ the result is obvious. If $x \in (0, 1)$ then we can easily see that $$\lim_{n \to \infty}n^{3}x^{n}(1 - x) = (1 - x)\lim_{n \to \infty}n^{3}x^{n}$$ To handle the limit of $a_{n} = n^{3}x^{n}$ I apply my favorite technique of ratio test. We have $$\lim_{n \to \infty}\frac{a_{n + 1}}{a_{n}} = \lim_{n \to \infty}\left(\frac{n + 1}{n}\right)^{3}x = x < 1$$ Hence the series $\sum a_{n}$ is convergent and therefore its $n$th term $a_{n}$ tends to zero.

If one does not have idea of series convergence and their tests then one can directly show that if $0 < \lim_{n \to \infty}a_{n + 1}/a_{n} = l < 1$ then $\lim_{n \to \infty} a_{n} = 0$. Clearly we can take a number $c$ such that $l < c < 1$ and then we know that there exists a positive integer $m$ such that $0< a_{n + 1}/a_{n} < c$ for all $n \geq m$ (use the $\epsilon$ definition of limit with $\epsilon = c - l > 0$). Then we get $$0 < \frac{a_{m + p}}{a_{m}} = \frac{a_{m + 1}}{a_{m}}\cdot \frac{a_{m + 2}}{a_{m + 1}}\cdots \frac{a_{m + p}}{a_{m + p - 1}} < c^{p}$$ Letting $p \to \infty$ and noting that $0 < c < 1$ we see that $\lim_{p \to \infty}a_{m + p}/a_{m} = 0$ and then $\lim_{p \to \infty}a_{m + p} = 0$ or what is the same as $\lim_{n \to \infty}a_{n} = 0$.

Another point which I gather from your question is about thinking of limit evaluation by "evaluation in parts". Like if two functions tend to limits $A$ and $B$ then their product tends to $AB$. This is a valid approach only when $A, B$ exist. Usually when one encounters a limit where evaluating part by part does not help (due to the fact that one of the parts does not tend to a finite limit) then we need to think a bit deeper.

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