Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$

Question

EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:

We do this using the epsilon-N definition of the limit of : $$\forall \varepsilon>0 ,\hspace{1mm} \exists N>0 \hspace{1mm}\text{s.t}\hspace{2mm} n\geq N \implies |a_{n}-L|<\varepsilon$$

Now, \begin{align} |a_{n}-L| & = \left| \frac{n^2+n-1}{n^2 + 2n +2}-1 \right| \\ &=\left| \frac{-n-3}{n^2 + 2n +2} \right| \end{align} We now consider the definition of $|x|$ for q quick moment, where we have that: $$|x|=\left\{ \begin{aligned} &x \hspace{2mm}\text{if} \hspace{2mm}x\geq0\\ -&x \hspace{2mm}\text{if} \hspace{2mm}x<0 \end{aligned} \right\} $$ Since we can split the absolute value into the numerator and denominator separately, we have that: $$|-n-3|=n+3 \hspace{2mm} \text{if} \hspace{2mm} -n-3<0 \iff -3<n$$ However, in our proof, we require $N>0$, thereby $n>0$, meaning that $|-n-3|=n+3$. A similar argument shows that $|n^2 +2n +2|=n^2+2n+2$ for $n>0$. Thereby, for $n>0$: \begin{align} |a_{n}-L| &=\frac{n+3}{n^2 + 2n +2} \\ &\leq \frac{n+3}{n^2+2n} \end{align} For $n\geq 3$, \begin{align}|a_n-L| &\leq \frac{n+3}{n^2+2n} \\ &\leq \frac{2n}{n^2+2n} \\ &= \frac{2}{n+2} \\ &\leq \frac{2}{n} < \varepsilon \end{align} Thus, to complete our proof, we choose $N=\max\{{\frac{2}{\varepsilon},3}\}$.

Does this look good?

Answer 1

HINT

As a suggestion, you do not need to split into two cases. Suppose that $n\geq n_{\varepsilon}$. Then one gets that \begin{align*} \left|\frac{n^{2} + n - 1}{n^{2} + 2n + 2} - 1\right| & = \frac{n + 3}{n^{2} + 2n + 2} = \frac{n + 3}{(n + 1)^{2} + 1} \leq \frac{3n + 3}{(n + 1)^{2}} \leq \frac{3}{n} \leq \frac{3}{n_{\varepsilon}} \end{align*}

Can you take it from here?

Answer 2

Do

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

Make use of the fact that this is each $1$ plus something that is getting smaller and smaller the bigger $n$ gets.

$\frac{1}{1+x}\approx 1-x$ in first order for small $x$.

So the division goes over into a simple binomial multiplication for larger $n$.

$(1+\frac{1}{n}-\frac{1}{n^2})(1-\frac{2}{n}-\frac{2}{n^2})$

The leading term is already the desired $1$ and the work is done without conducting the multiplication.

One can do $(1+\frac{1}{n})(1-\frac{2}{n})=1-\frac{1}{n}-\frac{2}{n^2}$

To show that the approximation towards one is for large natural $n$ dominated by $-\frac{1}{n}$.

The trick with $\frac{1}{1+x}\approx 1-x$ is usually or at least often introduced in courses facing this kind of tasks. An appropriate reference will be a plus in the answer.

Representate with a value table:

or the corresponding plot: