How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?

I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.

I started with our conergence definition, i.e. $\lvert a_n - L \rvert < \epsilon$

So $\lvert \frac{\sqrt {n^2 +2}}{4n+1} - \frac {1}{4} \rvert$ simplifies to $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}$

Now $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ is simplified to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$ Then I would square everything to remove the square root and simplify fractions but I end up with $n> \sqrt{\frac{1}{8(\epsilon^2-\frac{1}{16}}}$

We can't assume $\epsilon > \frac{1}{4}$ so somewhere I went wrong. Any help would be appreciated.

  • $\begingroup$ Are you set on using $\epsilon$? The problem is significantly more straightforward if you factor an $n$ from the numerator and denominator, as this leads to a ratio of two well-behaved functions with finite limits. $\endgroup$ – davidlowryduda Sep 7 '17 at 16:25
  • $\begingroup$ How do you go from $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4} < \epsilon$ to $\frac {4\sqrt {n^2 +2}}{16n} < \epsilon$? $\endgroup$ – José Carlos Santos Sep 7 '17 at 16:26
  • $\begingroup$ @JoséCarlosSantos I'm stating that since $\frac {4\sqrt{n^2+2}-4n-1}{16n+4} < \epsilon$ then $\frac{4\sqrt{n^2+2}}{16n}$ will also be less than epsilon since it will be a smaller fraction. $\endgroup$ – K Math Sep 7 '17 at 16:42
  • $\begingroup$ @mixedmath I am following the guidelines of the convergence proof. I stated the definition in the question and I also prove that $n \geqslant M$ I understand this isn't a good explanation of why but I need to keep the $\epsilon$ $\endgroup$ – K Math Sep 7 '17 at 16:46
  • $\begingroup$ @KellyR Really? Take $n=1$. Then $\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\simeq.0964102$ and $\frac {4\sqrt {n^2 +2}}{16n}\simeq0.433013$ . $\endgroup$ – José Carlos Santos Sep 7 '17 at 16:46

For one, we have $$ \frac{\sqrt{n^2+2}}{4n+1} \le \frac{\sqrt{n^2+2}}{4n} = \sqrt{\frac{1}{16}+\frac{1}{8n}} \le \frac{1}{4}+ \sqrt{\frac{1}{8n}}, $$ and we also have $$ \frac{\sqrt{n^2+2}}{4n+1} \ge \frac{n}{4n+1} = \frac{1}{4 + 1/n}. $$ As $n\to\infty$, both of these tend to $1/4$. By the squeeze theorem, we obtain the desired result.

If you need a more formal proof, you can go the extra mile like this:

Let $\epsilon>0$ be arbitrary. By the Archimedean property of $\Bbb R$, there exists $N_1\in\Bbb N$ such that if $n\ge N_1$, then $\sqrt{\dfrac{1}{8n}}<\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4}+\sqrt{\dfrac{1}{8n}} = \dfrac{1}{4}$.

Now consider the difference $$ \left|\frac{1}{4}-\frac{1}{4+1/n}\right| = \frac{1}{4}\cdot\frac{1}{4n+1}. $$ Again, by the Archimedean property of $\Bbb R$, there exists $N_2\in\Bbb N$ such that if $n\ge N_2$, then $\frac{1}{4}\cdot\frac{1}{4n+1} <\epsilon$. Thus $\lim_{n\to\infty}\dfrac{1}{4+1/n} = \dfrac{1}{4}$. By the squeeze theorem, the claim is proved.

  • $\begingroup$ I need a solution for n in the type of proof I am constructing. $\endgroup$ – K Math Sep 7 '17 at 17:33
  • $\begingroup$ @KellyR I am not sure what that means, but I updated my answer to include a more detailed proof that uses the Archimedean property of the reals. $\endgroup$ – Alex Ortiz Sep 7 '17 at 19:18

\begin{align} \lim_{n \rightarrow \infty} \dfrac{\sqrt{n^{2} + 2}}{4n + 1} = \lim_{n \rightarrow \infty} \dfrac{\frac{1}{n}\sqrt{n^{2} + 2}}{\frac{1}{n}(4n + 1)} = \lim_{n \rightarrow \infty} \dfrac{\sqrt{1 + \frac{2}{n^{2}}}}{\left(4 + \frac{1}{n} \right)} = \dfrac{1}{4} \end{align}


Let $\epsilon>0$

$$\left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right|\leq\frac{4n+8-4n-1}{16n+4}=\frac{7}{16n+4} \leq \frac{7}{16n}$$

We have that $\frac{7}{16n} \to 0$

Thus exists $n_0 \in \mathbb{N}$ such that $\frac{7}{16n}< \epsilon, \forall n \geq n_0$

So $\frac{1}{n}<\frac{16\epsilon}{7} \Rightarrow n> \frac{7}{16\epsilon}$

Take $n_0=[\frac{7}{16\epsilon}]+1$ and we have that $$\forall n\geq n_0=[\frac{7}{16\epsilon}]+1 \Rightarrow \frac{7}{16n}<\epsilon \Rightarrow \left|\frac {4\sqrt {n^2 +2} -4n-1}{16n+4}\right| < \epsilon $$

Note that $[x]$ is the integer part of $x$

  • $\begingroup$ I need a solution for n in the type of proof I am writing $\endgroup$ – K Math Sep 7 '17 at 17:33
  • $\begingroup$ @KellyR see my edits $\endgroup$ – Marios Gretsas Sep 7 '17 at 18:17

$\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14 $

$\begin{array}\\ \dfrac{\sqrt {n^2 +2}}{4n+1} &\gt \dfrac{n}{4n+1}\\ &= \dfrac{n+1/4-1/4}{4n+1}\\ &=\dfrac14- \dfrac{1}{4(4n+1)}\\ \end{array} $

so $\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14 \gt - \dfrac{1}{4(4n+1)} $.

Since $(n+1/n)^2 =n^2+2+1/n^2 \gt n^2+2 $, $\sqrt{n^2+2} \lt n+1/n $ so

$\begin{array}\\ \dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14 &\lt \dfrac{n+1/n}{4n+1}-\dfrac14\\ &= \dfrac{4n+4/n-(4n+1)}{4(4n+1)}\\ &= \dfrac{4/n-1}{4(4n+1)}\\ &= \dfrac{1}{n(4n+1)}-\dfrac{1}{4(4n+1)}\\ &\lt 0 \qquad\text{for } n > 4\\ \end{array} $

Therefore, for $n > 4$, $|\dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14| \lt \dfrac{1}{4(4n+1)} $.

By choosing $\dfrac{1}{4(4n+1)} \lt \epsilon $, or $n \gt \dfrac14(\dfrac1{4\epsilon}-1) $, the difference is less then $\epsilon$.

Two notes:

Choosing $n \gt \dfrac1{16\epsilon}$ is sufficient.

From the above, $0 \lt \dfrac{\sqrt {n^2 +2}}{4n+1}-\dfrac14+\dfrac{1}{4(4n+1)} \lt \dfrac{1}{n(4n+1)} $.


If you can use standard limit theorems about continuous functions, factor $n$ out of the numerator and the denominator (using fractions in both top and bottom), cancel them, etc.


Update If you really need to use a complete $\epsilon$-definition type proof, then using this method will only require that you additionally prove $\lim_{n\to\infty} \frac{n}{4n+1} = \frac{1}{4}$, which is considerably easier. (Just note that $\left|\frac{n}{4n+1}-\frac{1}{4}\right| = \frac{1}{4(4n+1)}$)

We will bound the terms eventually as follows: For any $\epsilon > 0$ there exists some $N$ so that for all $n\geq N$: $$n^2+2 < n^2(1+\epsilon)$$ (i.e., if $2 < N^2\epsilon$), also clearly: $$(1-\epsilon)n^2 < n^2+2$$

Putting these facts together, for $n\geq N$

$$\sqrt{1-\epsilon}\frac{n}{4n+1} = \frac{\sqrt{n^2(1-\epsilon)}}{4n+1} < \frac{\sqrt{n^2+2}}{4n+1} < \frac{\sqrt{n^2(1+\epsilon)}}{4n+1} = \sqrt{1+\epsilon}\frac{n}{4n+1} $$

Now taking limits we have: $$ \lim_{n\to\infty} \sqrt{1-\epsilon}\frac{n}{4n+1} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \lim_{n\to\infty} \sqrt{1+\epsilon}\frac{n}{4n+1} $$ so $$\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \frac{\sqrt{1+\epsilon}}{4}$$

as this holds for all $\epsilon > 0$, let $\epsilon \to 0^+$: $$\frac{1}{4} = \lim_{\epsilon\to 0^+}\frac{\sqrt{1-\epsilon}}{4} \leq \lim_{n\to\infty} \frac{n^2+2}{4n+1} \leq \lim_{\epsilon\to 0^+}\frac{\sqrt{1+\epsilon}}{4} = \frac{1}{4}$$ so the result follows from the sandwich theorem.


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