Using the $\varepsilon − N$ definition of the limit, prove that $\lim \limits_{n\to\infty} \frac{(n^2 + 1)}{ (n^2 + 2)} = 1$.

Question

Using the $ε − N$ definition of the limit, prove that $\displaystyle\lim \limits_{n\to\infty} \frac{(n^2 + 1)}{ (n^2 + 2)} = 1$.

In other words, given $\varepsilon> 0$, find explicitly a natural number $N$ which satisfies the statement in the definition of the limit.

Answer 1

Let $\varepsilon > 0$.

Define $$N = \lfloor \sqrt{\frac{1}{\varepsilon}-2}\rfloor +1$$

For all $n \geq N$, you have then $$n \geq \sqrt{\frac{1}{\varepsilon}-2}$$ so $$n^2+2 \geq \frac{1}{\varepsilon}$$

so $$\frac{1}{n^2+2} \leq \varepsilon$$

so $$\left| \frac{n^2+1}{n^2+2} - 1 \right| \leq \varepsilon$$

By the definition of the limit, this shows that $$\frac{n^2+1}{n^2+2} \rightarrow 1 $$

Answer 2

First of all rewrite the expression: $a_n := \frac{n^2+1}{n^2+2} = 1-\frac{1}{n^2+2}$.

Choose $N \in \mathbb{N}$ and let $n \geq N$. Then we have:

$ \Big | a_n -1 \Big | = \frac{1}{n^2+2} \leq \frac{1}{n^2} \leq \frac{1}{N^2}$

Do you have an idea how to choose $N \in \mathbb{N}$ from this inequality?

Answer 3

$$|a_n-1|={1\over n^2+2}$$therefore if $n>{1\over \sqrt{\epsilon}}$ we obtain $$n^2+2>{1\over \epsilon}+2>{1\over \epsilon}$$and $$|a_n-1|<\epsilon$$ hence the result.