# Thread: cannot find limit of a sequence because of a non factoring fraction

1. ## cannot find limit of a sequence because of a non factoring fraction

Hello,

I am trying to find the limit of this sequence:

$\sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0$

but I cannot factorize the fraction at all.

Any idea is very much appreciated!

Thank you all!

2. ## Re: cannot find limit of a sequence because of a non factoring fraction

Originally Posted by Melsi
I am trying to find the limit of this sequence:
$\sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}, k,l>0$
You should know this theorem:
If $0 then $\left( {\sqrt[n]{{a^n + b^n }}} \right) \to b$.

3. ## Re: cannot find limit of a sequence because of a non factoring fraction

Hello,

no it is the very first time I see this theorem, I will check it out tomorrow (02:00 at midnight now) and come back with a full solution!

Very kind of you!
Thank you very much!

4. ## Re: cannot find limit of a sequence because of a non factoring fraction

Here is the proof of that theorem.
$b = \sqrt[n]{{b^n }} \leqslant{\color{blue} \sqrt[n]{{a^n + b^n }}} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2} \to b$

5. ## Solved!

The proof... that's a good one, cool! Now it can be solved..

the initial formula of the general term could be written like this:

$\sqrt[n]{\frac{k^{n}+l^{n}}{k^{2n}+l^{2n}}}=\frac{(k^{n}+l ^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}$

and the limit like this:

$\lim_{n \to \infty}{a_{n}=\lim_{n\to \infty }\frac{(k^{n}+l^{n})^{1/n}}{(k^{2n}+l^{2n})^{1/n}}=\frac{\lim_{n\to \infty }(k^{n}+l^{n})^{1/n}}{\lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}}$

if we apply the theorem for the numerator we have:

$l\leq \lim_{n\to \infty }(k^{n}+l^{n})^{1/n}\leq \lim_{n\to \infty }(l+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{n}+l^{n})^{1/n} = l$

and for the denominator it goes like this:

$l^{2}\leq \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n}\leq \lim_{n\to \infty }(l^{2}+\sqrt[n]{2})\Leftrightarrow \lim_{n\to \infty }(k^{2n}+l^{2n})^{1/n} = l^{2}$

Finally having found those two limits we can write:

$\lim_{n \to \infty}a_{n}=\frac{l}{l^{2}}=\frac{1}{l}$

and all these supposing that $a\leq b$ and $a,b> 0$ this means that if $b\leq a$ it will go the other way around, the limit will be
$\frac{1}{k}$

Thank you once again, for your valuable help!