Limit of a
- Thread starter dome123
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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?
[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?
Cubist
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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]
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Cubist
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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 + ...
Cubist
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Not quite. Using the a and b notation:-
[math] (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} [/math]
where
[math] a= \sqrt{n+\sqrt{n+\sqrt{n}}} [/math][math] b=\sqrt{n}[/math]
If we concentrate on the numerator:-
[math] a^2 - b^2 [/math][math] = \left(n+\sqrt{n+\sqrt{n}}\right) - \left(n\right) [/math][math] = n+\sqrt{n+\sqrt{n}} - n [/math][math] = \sqrt{n+\sqrt{n}} + \left(n-n\right) [/math][math] = \sqrt{n+\sqrt{n}} [/math]
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.
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.
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