# Thread: How do you find the limit of these sequences?

1. ## How do you find the limit of these sequences?

a) (2n^2+n+sqrt(n)-3)/(n^2 -5n+7) as n tends to infinity
b) (1+2+3+...+n)/n^2 as n tends to infinity

2. For the first, we may divide the numerator and denominator by $\displaystyle n^2$. For the second, we may use the formula

$\displaystyle 1+2+\cdots+n=\frac{n(n+1)}{2}.$

3. Originally Posted by amm345
a) (2n^2+n+sqrt(n)-3)/(n^2 -5n+7) as n tends to infinity
b) (1+2+3+...+n)/n^2 as n tends to infinity

$\displaystyle \frac{2n^2+n+\sqrt{n}-3}{n^2-5n+7}=\frac{2n^2+n+\sqrt{n}-3}{n^2-5n+7}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}$

$\displaystyle =\frac{2+\frac{1}{n}+\frac{1}{n^{3/2}}-\frac{3}{n^2}}{1-\frac{5}{n}+\frac{7}{n^2}}$

and you should be able to see now what the limit is....

Always multiply by 1 in the special form of 1 over the highest term of the fraction, in fact, this idea leads to a rule about limits to infinity

The highest power term always dominates, so if its on top then we get infinity, if it's on bottom we get 0, if the degree of top and bottom are equal, then the limit is the ratio of coefficients between the highest degree terms as is the case here

If you want to see this argument rigorously, let me know

4. I think I understand it and that was the limit I got the first time I tried, I just wanted to be sure, but if you don't mind seeing the rigorous version would help!

5. Originally Posted by amm345
I think I understand it and that was the limit I got the first time I tried, I just wanted to be sure, but if you don't mind seeing the rigorous version would help!

Alright, let's only deal with polynomials here...

Say we have $\displaystyle \lim_{n\rightarrow\infty}\frac{a_0n^{p_0}+a_1n^{p_ 1}+...+a_n}{b_0n^{q_0}+b_1n^{q_1}+...+b_n}$

Now we might as well assume that $\displaystyle p_0>p_i$ for all other$\displaystyle i$ since if it isn't, we just rearrange the expression on top to make it that way.... same on the bottom

So now, 1 of 3 things can happen, either $\displaystyle p_0=q_0,$ $\displaystyle p_0>q_0$ or $\displaystyle p_0<q_0$

So case 1, what if $\displaystyle p_0=q_0$?

$\displaystyle \frac{a_0n^{p_0}+a_1n^{p_1}+...+a_n}{b_0n^{q_0}+b_ 1n^{q_1}+...+b_n}=\frac{a_0n^{p_0}+a_1n^{p_1}+...+ a_n}{b_0n^{q_0}+b_1n^{q_1}+...+b_n}\cdot \frac{\frac{1}{n^{p_0}}}{\frac{1}{n^{p_0}}}$

$\displaystyle =\frac{a_0+a_1n^{p_1-p_0}+...+\frac{a_n}{n^{p_0}}}{b_0+b_1n^{q_1-p_0}+...+\frac{b_n}{n^{p_0}}}$

Notice since we assume $\displaystyle p_0=q_0$ the cancellation happens on both the top and bottom. Also, since we assume both $\displaystyle p_0$ and $\displaystyle q_0$ are the largest exponents, everything left has a negative exponent, which means it is of the form $\displaystyle c_i\frac{1}{n^y}$ where $\displaystyle c_i$ and $\displaystyle y$ are just 2 numbers. So the limit as n goes to infinity is zero in all those terms. So the limit is the only thing left which is $\displaystyle \frac{a_0}{b_0}$

Now let's assume $\displaystyle p_0>q_0$

Then we do the same thing but we multiply top and bottom by $\displaystyle \frac{\frac{1}{n^{q_0}}}{\frac{1}{n^{q_0}}}$

This will cancel everything on the bottom or give it a negative exponent and since $\displaystyle p_0>q_0$ at least the first term of the numerator will still have a positive exponent, so we have something of the form $\displaystyle c_in^y$
Which clearly goes to infinity

Now let's assume $\displaystyle p_0<q_0$

Then we do the same thing but we multiply top and bottom by $\displaystyle \frac{\frac{1}{n^{p_0}}}{\frac{1}{n^{p_0}}}$

So everything on the top cancels or has a negative exponent, and at least the first term of the denominator has a positive exponent so we have something of the form $\displaystyle c_i\frac{1}{n^y}$ which clearly goes to zero

I can spell it fully out like the first example if needed, but I think it should be clear