Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to evaluate $\lim\limits_{n\to\infty}(n^3\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}))$?

My though is, that since $\lim\limits_{n\to\infty}n^3=\infty$, the limit must be $\infty$, if the right hand side is greater than or equal to $1$. Otherwise it's a fraction, and there might be many possible solutions.

However, I'm far for being confident about the solution. What am I missing here? What's the "trick"?

share|cite|improve this question
A basic strategy when you have limits involving radicals is to multiply by the conjugate over the conjugate. – Hakim May 27 '14 at 20:49
up vote 2 down vote accepted

Hint: \begin{align}\\&\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2} \\\\=&\dfrac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}} \\\\=&\dfrac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}} \\\\=&\dfrac{n^4+1-n^4}{\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)\cdot\left(\sqrt{n^4+1}+n^2\right)}\end{align}

share|cite|improve this answer
This hint was helpful, thanks! Also, if somebody is not familiar with the conjugate over conjugate multiplication, this link will be helpful:… – kdani May 27 '14 at 21:03

Recall that by the Taylor series we have $$(1+x)^\alpha\sim_01+\alpha x$$ so we apply it:

$$\sqrt{n^2+\sqrt{n^4+1}}=n\sqrt{1+\sqrt{1+\frac1{n^4}}}\sim_\infty n\left(1+1+\frac{1}{2n^4}\right)^{1/2}\\=n\sqrt2\left(1+\frac{1}{4n^4}\right)^{1/2}\sim_\infty n\sqrt2\left(1+\frac{1}{8n^4}\right)$$ hence we see that


share|cite|improve this answer

Apply conjugation twice to get the numerator to be $n^3$, then:$L \sim \dfrac{n^3}{2\sqrt{2}n\times 2n^2} \to \dfrac{1}{4\sqrt{2}}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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