Find $\lim_{n\to\infty}\frac{n(n-1)/2}{n^3}$

Question

Find $$\lim_{n\to\infty}\frac{n(n-1)/2}{n^3}$$

I'm trying to use L'Hospitals rule and I am stuck. So far I have:

$$\lim_{n\to\infty}\frac{n(n-1)/2}{n^3}=\lim_{n\to\infty}\frac{2(n-1)-(2n-1)(0)}{(n^3)^2}=\lim_{n\to\infty}\frac{2(2n-1)-(2n-1)(0)}{n^6}$$

Answer 1

HINT:

• L'Hopital's rule states that under ___ conditions, $$\lim\frac fg=\lim\frac{f'}{g'}\ne\lim\left(\frac fg\right)'$$

• So you get $$\frac12\lim_{n\to\infty}\frac{n^2-n}{n^3}=\frac12\lim_{n\to\infty}\frac{2n-1}{n^2}=\cdots$$

Answer 2

There are some arithmetic errors in your calculus. Differentiating the numerator and denominator once respectively gives $n-1/2,\,3n^2$. Doing it again gives $1,\,6n$. The ratio has limit $0$. You can also solve the problem without L'Hopital's rule; for example, $x=1/n$ gives $\lim_{x\to 0^+}\frac{x(1-x)}{2}$.

Answer 3

Hint You can also do it without l hospital's rule. n(n-1)/2(n^3)=(n-1)/2(n^2)=(1/2n)-(1/2(n^2)) From here you can easily put the limits It can also be solved with l hospital's rule by not making differentiation complex.

Answer 4

We have that

$$\frac{n(n-1)/2}{n^3}=\frac12\frac {n^2}{n^3}(1-1/n)=\frac12\frac {1}{n}(1-1/n)\to \frac12\cdot 0\cdot1=0$$

Answer 5

Let $n > 2$.

$0 < \dfrac{n(n-1)}{2n^3} =$

$\dfrac{n-1}{2n^2} < \dfrac {n-1}{2n(n-1)}=$

$(1/2)\dfrac{1}{n}.$

Take the limit $n \rightarrow \infty$ using squeeze theorem.

Answer 6

Just divide by $n^3$:

$$\displaystyle\lim_{n\to\infty}\frac{\frac{n(n-1)}{2n^3}}{\frac{n^3}{n^3}}=\displaystyle\lim_{n\to\infty}\frac{\frac{n^2-1}{2n^3}}{1}=\lim_{n\to\infty}\Big(\frac{1}{2n}-\frac{1}{2n^2}\Big)=0$$

If you had had $n^4$ instead of $n^3$, then you will have to divide by $n^4$ and so on.