Limit of a

[dome123]

Hello,

I have a problem solving a limit of a number series which is given with the n-th term:

I appreciate all the help as this problem doesn't seem easy(for me at least).

 

[MarkFL]

I think I would look at writing:

[MATH]\sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}\cdot\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}=\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}[/MATH]
Now, divide numerator and denominator by $\sqrt{n}$...what do you get?

 

[dome123]

I have done exactly that, but I divided with n instead of sqrt(n).
Thanks for the feedback, I'll try to solve it that way.

Edit: Are you missing an n in the numerator ---> sqrt(n+sqrt(n)) - n

 

[Cubist]

Edit: Are you missing an n in the numerator ---> sqrt(n+sqrt(n)) - n

I think MarkFL's numerator was correct, but the LHS probably needed some brakets:

[MATH]\left( \sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}\right) \cdot\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}=\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}[/MATH]

 

[MarkFL]

I think MarkFL's numerator was correct, but the LHS probably needed some brakets:

[MATH]\left( \sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}\right) \cdot\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}=\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n+\sqrt{n+\sqrt{n}}}+\sqrt{n}}[/MATH]

Yes, that was sloppy of me...thanks for the clarification.

 

[dome123]

I can't see how the original -sqrt(n) disappears.
Isn't (a-b)(a+b) = a^2- b^2 ?

 

[Cubist]

Isn't (a-b)(a+b) = a^2- b^2 ?

Yes. You'll see what is happening if you think about what a and b are.
In this case a=sqrt(n + xxx) and b=sqrt(n) therefore b^2=n and a^2=n + ...

 

[dome123]

Yes. You'll see what is happening if you think about what a and b are.
In this case a=sqrt(n + xxx) and b=sqrt(n) therefore b^2=n and a^2=n + ...

Wouldn't it look like this then :

 

[Cubist]

Wouldn't it look like this then :View attachment 16736

Not quite. Using the a and b notation:-

$$(a-b)\left(\frac{a+b}{a+b}\right) \to \frac{\left(a-b \right)\left(a+b\right)}{a+b} \to \frac{\left(a^2 - b^2\right)}{a+b}$$
where

$$a= \sqrt{n+\sqrt{n+\sqrt{n}}}$$$$b=\sqrt{n}$$
If we concentrate on the numerator:-

$$a^2 - b^2$$$$= \left(n+\sqrt{n+\sqrt{n}}\right) - \left(n\right)$$$$= n+\sqrt{n+\sqrt{n}} - n$$$$= \sqrt{n+\sqrt{n}} + \left(n-n\right)$$$$= \sqrt{n+\sqrt{n}}$$
Does that make sense? If you have a different way of doing this that leaves an extra "n" then please post a picture of your working and I'll see if I can spot a mistake.

 

[lookagain]

Hello,

I have a problem solving a limit of a number series which is given with the n-th term:

dome123,

this first post has n approaching negative infinity. It should have been n approaching infinity, from the start.

You could let $\displaystyle \ n \ = \ x^2$.

$\displaystyle \displaystyle\lim_{n \to \ \infty} \bigg(\sqrt{x^2 \ + \ \sqrt{x^2 \ + \ x \ }} \ - \ x\bigg)$

Multiply by the conjugate in each place, similarly in post #5, to get

$\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{\sqrt{x^2 \ + \ x}}{\sqrt{x^2 \ + \ \sqrt{x^2 \ + \ x \ }} \ + \ x}$


This will reduce the number of square root signs from six in post #4 down to three.

$\displaystyle \sqrt{x^2} \ = \ |x|, \ \$ but you can ignore that and write $\displaystyle \ \sqrt{x^2} \ = \ x \ \$as you factor it out from the numerator and
the denominator, because $\displaystyle \ x \to \ \infty$.


$\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{\sqrt{x^2( 1 \ + \ \tfrac{1}{x})}}{\sqrt{x^2 \ + \ \sqrt{x^2( 1 \ + \ \tfrac{1}{x})} } \ + \ x}$

$\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{\sqrt{x^2 \ + \ x\sqrt{ 1 \ + \ \tfrac{1}{x}} } \ + \ x}$

$\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{\sqrt{x^2( 1 \ + \ \tfrac{1}{x}\sqrt{ 1 \ + \ \tfrac{1}{x}}) } \ + \ x}$

$\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{x\sqrt{ 1 \ + \ \tfrac{1}{x}\sqrt{ 1 \ + \ \tfrac{1}{x}} } \ + \ x}$


Continue factoring out the x variables, divide them/cancel them out, and continue.

 

[dome123]

I got it now, thanks for the help guys.