1
$\begingroup$

This question is spurred by my discussion here: Find the value of : $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$

My question is how do you prove what @Ant is claiming? It looks like a very useful simplification when the number of nested radical signs grow big. So I'm looking for a proof of:

$$\lim_{n \to \infty}{\sqrt{n+\sqrt{n}}} = \lim_{n \to \infty}{\sqrt{n}}$$

or at least a proof that one of the expressions can be substituted for the other in limit calculations.

$\endgroup$
  • 1
    $\begingroup$ Both $\sqrt{n+\sqrt{n}}$ and $\sqrt{n}$ tend to infinity as $n$ tends to infinity. $\endgroup$ – Shaun Mar 5 '17 at 17:59
  • 1
    $\begingroup$ Requiring that one expression "can be substituted for the other" is a much stronger (not weaker) condition than both expressions having equal limits. $\endgroup$ – hardmath Mar 5 '17 at 18:07
  • $\begingroup$ Maybe you can view this as an application of the squeeze theorem if you have heard of it. $\endgroup$ – mathreadler Mar 5 '17 at 18:20
  • $\begingroup$ Sorry I'm not very good at calculus. You have to try and understand what I really mean. $\endgroup$ – Björn Lindqvist Mar 5 '17 at 18:21
  • 1
    $\begingroup$ kingW3's answer is the correct one to your question (namely what you need for the reasoning in the linked post). Note that what you need is not "can always be substituted" which is never true, but rather "have ratio approaching $1$", which is in this case true. $\endgroup$ – user21820 Mar 11 '17 at 11:27
2
$\begingroup$

Assuming you want to prove that $$\lim_{n\to\infty}\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n}}=1$$ This follows by putting everything into one square root. $$\lim_{n\to\infty}\sqrt{\frac{n+\sqrt{n}}{n}}=\lim_{n\to\infty}\sqrt{1+\frac{1}{\sqrt{n}}}=1$$ Another way to show that is to show $$\sqrt{n}\leq\sqrt{n+\sqrt{n}}\leq\sqrt{n}+\frac12$$

$\endgroup$
4
$\begingroup$

Are you happy that $$\lim_{n \to \infty} \sqrt n = \infty?$$

If so you just need to note that

$$\sqrt n \le \sqrt{n + \sqrt n}$$

and then take the limit as $n \to \infty$.

$\endgroup$
1
$\begingroup$

Note however that: $$\lim_{n\to\infty} \left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right)\ne 0$$

Multiply and divide by the rational conjugate:

$$ \require{cancel} \begin{align} \lim_{n\to\infty} \left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right) & = \lim_{n\to\infty} \frac{\left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right)\left(\sqrt{n+\sqrt{n}}+ \sqrt{n}\right)}{\sqrt{n+\sqrt{n}}+ \sqrt{n}} \\ & = \lim_{n\to\infty} \frac{\cancel{n}+\sqrt{n} - \cancel{n}}{\sqrt{n+\sqrt{n}}+ \sqrt{n}} \\ & = \lim_{n\to\infty} \frac{1}{\sqrt{1 + \cfrac{1}{\sqrt{n}}}+ 1} \\ & = \frac{1}{2} \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.