1
$\begingroup$

Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.

We have $$\begin{align} \left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2n^2-2-2n^2-3}{2(2n^2+3)}\right| \\ &= \left|\frac{-5}{2(2n^2+3)}\right|\\ &= \frac{5}{2(2n^2+3)} \\ &<\frac{5}{4n^2}, \end{align}$$

$$\frac{5}{4n^2}<\epsilon \iff \frac{1}{n^2}<\frac{4 \epsilon}{5} \iff n>\sqrt{\frac{5}{4\epsilon}}$$

We choose $n_0=\left[\sqrt{\frac{5}{4\epsilon}} \right]+1$, Then $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$.

Let $(x_n)=\frac{1}{\ln(n+1)}$ for $n \in \mathbb{N}$.

a) Use the definition of the limit to show that $\lim(x_n)=0$.

$|\frac{1}{\ln(n+1)}-0|=\frac{1}{\ln(n+1)}<\epsilon \Leftrightarrow ln(n+1) > \epsilon \Leftrightarrow n> e^{\epsilon} -1$

We choose $n_0=\left[ e^\epsilon -1 \right]+1$, Then $\lim(x_n)=0$.

b) Find specific value of $n_0 (\epsilon)$ as required in definition of limit for $\epsilon=\frac{1}{2}$.

$n_0=\left[\sqrt{e}-1\right]+1$

Is that true, please?

$\endgroup$
4
  • 1
    $\begingroup$ In your second exercise, you should have $\ln(n+1)>1/\epsilon$. $\endgroup$ Feb 3 '19 at 1:25
  • 1
    $\begingroup$ You are only showing the work in finding an $n_0$. The actual proof should look like "Let $\epsilon > 0$, and let $n_0 =$ ____. Then if $n>n_0$ .... work ... $|a_n - L| < \epsilon$. QED" $\endgroup$
    – David P
    Feb 3 '19 at 1:26
  • $\begingroup$ @JohnWaylandBales You are right. Thank you so much. $\endgroup$
    – Dima
    Feb 3 '19 at 1:35
  • $\begingroup$ @DavidPeterson Thank you so much. $\endgroup$
    – Dima
    Feb 3 '19 at 1:35
1
$\begingroup$

If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:

Prove: $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$

Proof: Let $\epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right|<\epsilon$

Then proceed with the steps which you have given.

$\endgroup$
1
  • $\begingroup$ Thank you so much. $\endgroup$
    – Dima
    Feb 3 '19 at 1:57
1
$\begingroup$

For a) We have $\lim_{n\rightarrow \infty} \frac{n^2-1}{2n^2+3}=\lim_{n\rightarrow \infty} \frac{n^2(1-\frac{1}{n^2})}{n^2(2+\frac{3}{n^2})}= \lim_{n\rightarrow \infty} \frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}=\frac{1}{2}$

For b) basically the same,meaning $ln$ is monotone so for $n\rightarrow \infty$ it follows that $\ln(n)\rightarrow \infty$

$\endgroup$
1
  • $\begingroup$ Thanks a lot, sorry I forget to write "by the definition". $\endgroup$
    – Dima
    Feb 3 '19 at 1:34
1
$\begingroup$

Your answer is correct. However it seems you may have overcomplicated it.

\begin{align} \lim_{n\to \infty}\left(\frac{n^2-1}{2n^2+3}\right) & = \lim_{n\to \infty}\left(\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}\right)\\ \end{align}

Now use the fact that $\lim_{n\to \infty}\left(\frac{1}{n^k}\right)=0$, where $k$ is any positive integer.

Hence \begin{align} \lim_{n\to \infty}\left(\frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}\right) & = \lim_{n\to \infty}\left(\frac{1-0}{2+0}\right)\\ &= \frac{1}{2} \end{align}

$\endgroup$
1
  • $\begingroup$ Thanks a lot, sorry I forget to write "by the definition". $\endgroup$
    – Dima
    Feb 3 '19 at 1:34

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.