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?
Best Answer
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.