# Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$

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 $$\epsilon=\frac{1}{2}$$.

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

• In your second exercise, you should have $\ln(n+1)>1/\epsilon$. Feb 3 '19 at 1:25
• 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" Feb 3 '19 at 1:26
• @JohnWaylandBales You are right. Thank you so much.
– Dima
Feb 3 '19 at 1:35
• @DavidPeterson Thank you so much.
– Dima
Feb 3 '19 at 1:35

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.

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

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 $$\ln(n)\rightarrow \infty$$

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

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}

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