Is my proof of $\lim _{n\to \infty \:}\left(\frac{n^2-1}{n+1}\right)=\infty $ accurate?

Question

prove $$\lim _{n\to \infty \:}\left(\frac{n^2-1}{n+1}\right)=\infty $$

My proof: For any $M ∈ ℝ$ there exists $N ∈ ℕ$ such that n>N (n is sufficiently large) and so choose $a_n$>M

$\left(\frac{n^2-1}{n+1}\right)$ > M

$\frac{\frac{n^2}{n}-\frac{1}{n}}{\frac{n}{n}+\frac{1}{n}}$ >M

n>M

This shows for any M I can find $a_n$ where n>M, therefore the sequence approaches infinity.

Does this make sense??

Answer 1

To prove by the definition that

$$ \lim _{n\to \infty \:}\left(\frac{n^2-1}{n+1}\right)=\infty $$

you must show that if $M>0$ then there is an $n_0>0$ such that for any $n>n_0$, $\frac{n^2-1}{n+1}>M$.

Proof: Let $M>0$. Let $n_0=M+1$. Then if $n>n_0$

\begin{eqnarray} n-1&>&M\\ (n-1)\cdot\frac{n+1}{n+1}&>&M\\ \frac{n^2-1}{n+1}&>&M \end{eqnarray}

Answer 2

Factor with difference of squares to get $$\lim_{n\to\infty}\frac{n^2-1}{n+1}=\lim_{n\to\infty}\frac{(n-1)(n+1)}{n+1}=\lim_{n\to\infty}(n-1)=\infty$$

Answer 3

Simplify the limit first: $$\lim_{n\to ∞} \frac{n^2-1}{n+1} = \lim_{n\to ∞} \frac{(n-1)(n+1)}{n+1} = \lim_{n\to ∞} (n-1)$$

Hence proving $$\lim_{n\to ∞} \frac{n^2-1}{n+1} = ∞$$ is proving $$\lim_{n\to ∞} (n-1) = ∞$$

Answer 4

Remember that because:

$$(n-1)(n+1)=n^2-1$$we have that:

$$n\neq -1 \implies \frac{n^2-1}{n+1}=n-1$$

Since we are taking the limit as $n\to \infty$, the condition can be ignored so we just find:

$$\lim_{n\to\infty}(n-1)$$ which is trivially $\infty$