Thread: limit and dividing by n^(3/4) CORRECTION

1. limit and dividing by n^(3/4) CORRECTION

this is a simple question but i cant seem to get it.
I have to find

$\displaystyle [\lim_{n \to \infty} (n^{\frac{3}{4}})((n+1)^{\frac{1}{4}}- (n-1)^{\frac{1}{4}})$

i have got to

$\displaystyle [\lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}$

and know that i have to divide by n^(3/4) to get the answer 1/2 as n tend to infinity.
I keep making mistakes when dividing by this, can anyone help step by step?
Thanks

2. Originally Posted by TS1
this is a simple question but i cant seem to get it.
I have to find

$\displaystyle [\lim_{n \to \infty} (n^{\frac{3}{4}})((n+1)^{\frac{1}{4}}- (n-1)^{\frac{1}{4}})$

i have got to

$\displaystyle [\lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}$

and know that i have to divide by n^(3/4) to get the answer 1/2 as n tend to infinity.
I keep making mistakes when dividing by this, can anyone help step by step?
Thanks
Dear TS1,

Can you tell me the answers for,

$\displaystyle \frac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}=$ ?

$\displaystyle \frac{(n+1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}}{n^{\frac{3}{4}}}=$ ?

3. To the given problem, multiply and divide by
[(n+1)^1/4 + (n-1)^1/4][(n+1)^1/2 + (n-1)^1/2]
and simplify.

4. $\displaystyle \frac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}=$

(1+1/n)^(3/4)?

i cant do these though

$\displaystyle \frac{(n+1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}}{n^{\frac{3}{4}}}=$

$\displaystyle \frac{(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}}{n^{\frac{3}{4}}}=$

5. Your first simplification is correct.
Now as n tends to infinity 1/n tends to zero.

Similarly you can simplify the other two.

6. how do i do that when the powers are different?

7. Hello, TS1!

You are that close . . .

I have got to:
$\displaystyle \lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}$

I know that I have to divide by $\displaystyle n^{\frac{3}{4}}$ to get the answer $\displaystyle \tfrac{1}{2}.$ . . Right!

Divide top and bottom by $\displaystyle n^{\frac{3}{4}}$

$\displaystyle \frac{2}{\dfrac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4 }}} + \dfrac{(n+1)^{\frac{2}{4}}}{n^{\frac{2}{4}}} \dfrac{(n-1)^{\frac{1}{4}}}{n^{\frac{1}{4}}} + \dfrac{(n+1)^{\frac{1}{4}}}{n^{\frac{1}{4}}}\dfrac {(n-1)^{\frac{2}{4}}}{n^{\frac{2}{4}}} + \dfrac{(n-1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}}$

. . $\displaystyle =\; \frac{2} {\left(\dfrac{n+1}{n}\right)^{\frac{3}{4}} + \left(\dfrac{n+1}{n}\right)^{\frac{2}{4}} \left(\dfrac{n-1}{n}\right)^{\frac{1}{4}} + \left(\dfrac{n+1}{n}\right)^{\frac{1}{4}} \left(\dfrac{n-1}{n}\right)^{\frac{2}{4}} + \left(\dfrac{n-1}{n}\right)^{\frac{3}{4}}}$

. . $\displaystyle =\; \frac{2} {\left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{2}{4}} \left(1 - \dfrac{1}{n}\right)^{\frac{1}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{1}{4}}\left(1 - \dfrac{1}{n}\right)^{\frac{2}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}}}$

Then: . $\displaystyle \lim_{n\to\infty}\left[ \frac{2} {\left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{2}{4}} \left(1 - \dfrac{1}{n}\right)^{\frac{1}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{1}{4}}\left(1 - \dfrac{1}{n}\right)^{\frac{2}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}}}\right]$

. . . . $\displaystyle =\;\; \frac{2} {(1 + 0)^{\frac{3}{4}} + (1 + 0)^{\frac{2}{4}} (1 - 0)^{\frac{1}{4}} + (1 + 0)^{\frac{1}{4}}(1 - 0)^{\frac{2}{4}} + (1 + 0)^{\frac{3}{4}}}$

. . . . $\displaystyle =\;\;\frac{2}{1+ 1\!\cdot\!1 + 1\!\cdot\!1 + 1 } \;\;=\;\;\frac{2}{4} \;\;=\;\;\frac{1}{2}$