# evaluate the limit

#### Vali

##### New member
$$\displaystyle a_{0},a_{1},...,a_{k}$$ are real numbers and $$\displaystyle a_{0}+a_{1}+...+a_{k}=0$$
$$\displaystyle L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
$$\displaystyle L=?$$

I have no idea how to start, I think I should write difference of sqrt to find the limit.

#### Jomo

##### New member
$$\displaystyle a_{0},a_{1},...,a_{k}$$ are real numbers and $$\displaystyle a_{0}+a_{1}+...+a_{k}=0$$
$$\displaystyle L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
$$\displaystyle L=?$$

I have no idea how to start, I think I should write difference of sqrt to find the limit.
What can you say about $$\displaystyle L=\lim_{n\rightarrow \infty }(a_{r}\sqrt[3]{n+r})$$ for any r > 0? That is, what, if anything, is the difference between these limits with different r?

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#### Vali

##### New member
I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution.

$$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$$ so $$\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$$

$$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$

Now I group the terms and I have

$$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$$

I split in more limits, the constants $$\displaystyle a_{0},a_{1},...,a_{k}$$ go ahead the limits and the limits are 0 after some calculations.In the end $$\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$$

#### Jomo

##### New member
I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution.

$$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$$ so $$\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$$

$$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$

Now I group the terms and I have

$$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$$ On one hand we get [tex}\infty-\infty[/tex]

I split in more limits, the constants $$\displaystyle a_{0},a_{1},...,a_{k}$$ go ahead the limits and the limits are 0 after some calculations.In the end $$\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$$
So you are not using a[SUB]1 [/SUB]+ a[SUB]2[/SUB] + ... +a[SUB]k[/SUB] = 0. It could be possible that this is not needed. My main concern is that you think that inf - inf = 0. This is not true at all. Consider $$\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - ]$$. On one hand, you get $$\displaystyle \infty - \infty$$, while on the otherhand, n+k - n =k, so the $$\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - ]$$ = k, for some number k.

So in the end we see that some limits that give us $$\displaystyle \infty - \infty$$ can actually equal ANY NUMBER.

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#### pka

##### Active member
I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution
$$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$$ so $$\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$$ (1)
$$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$ (2)
Now I group the terms and I have
$$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$$ (3)
I think you, Vali, are correct in that $$\displaystyle \large\color{blue}{(1)\to(2)\to(3)}$$

Lets point out that:
\displaystyle \begin{align*}\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)&=\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)\left(\dfrac{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}{\sqrt[3]{(n+k)^2}\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right)\\&=\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right) \end{align*}

#### Vali

##### New member
So the limit is not 0 too ?
Because we have n at denominator

#### Jomo

##### New member
So the limit is not 0 too ?
Because we have n at denominator
Why do you think that? For example what does 1/n approach as n approaches infinity?

#### Vali

##### New member
I'm 99% sure it's 0.

#### pka

##### Active member
I think there's no difference because everytime L tends to infinity Meanwhile, I think I found a solution.
$$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$$ so $$\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$$
$$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
Now I group the terms and I have
$$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$$
I split in more limits, the constants $$\displaystyle a_{0},a_{1},...,a_{k}$$ go ahead the limits and the limits are 0 after some calculations.In the end $$\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$$
So the limit is not 0 too ?
Because we have n at denominator
I'm 99% sure it's 0.
Lets point out that:
\displaystyle \begin{align*}\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)&=\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right) \end{align*}
$$\displaystyle L=\displaystyle\lim_{n\rightarrow \infty }\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right)=0$$