2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

\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}

$\endgroup$
1
$\begingroup$

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 $x$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$\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)} $.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

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.