Question

Q: Find the limit of the following sequences:
(a) lim2+(1/n2), (b)lim(-1)n/(n+2)
(c)lim((√n-1)/(√n+1))
(d)lim((n+1)/(n(√n)))

Answer 1

Answer on Question #64526 – Math – Real Analysis

Question

Find the limit of the following sequences:

(a) $\lim \left(2 + \frac{1}{n^2}\right)$, (b) $\lim \frac{(-1)^n}{n+2}$,

(c) $\lim \frac{\sqrt{n} - 1}{\sqrt{n} + 1}$

(d) $\lim \frac{n + 1}{n\sqrt{n}}$

Solution

(a) $\lim_{n\to \infty}\left(2 + \frac{1}{n^2}\right) = \lim_{n\to \infty}2 + \lim_{n\to \infty}\frac{1}{n^2} = 2 + \lim_{n\to \infty}\frac{1}{n^2}.$

So if $\varepsilon > 0$ is given, then choosing $N > \frac{1}{\sqrt{\varepsilon}}$ we have $\frac{1}{n^2} < \varepsilon$, $n > N$.

Thus $\lim_{n\to \infty}\frac{1}{n^2} = 0$ and therefore $\lim_{n\to \infty}\left(2 + \frac{1}{n^2}\right) = 2$.

(b) $\lim_{n\to \infty}\frac{(-1)^n}{n + 2}$.

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

Since $\lim_{n\to \infty}\frac{-1}{n + 2} = \lim_{n\to \infty}\frac{1}{n + 2} = 0$, by Squeeze Theorem, $\lim_{n\to \infty}\frac{(-1)^n}{n + 2} = 0$.

(c) $\lim_{n\to \infty}\frac{\sqrt{n} - 1}{\sqrt{n} + 1}$.

$$\left| \frac {\sqrt {n} - 1}{\sqrt {n} + 1} - 1 \right| = \left| \frac {\sqrt {n} - 1 - (\sqrt {n} + 1)}{\sqrt {n} + 1} \right| = \frac {2}{\sqrt {n} + 1}$$

If $\varepsilon > 0$ is given, then choosing $N > 1 + \left(\frac{2}{\varepsilon}\right)^2$ we have $\left|\frac{\sqrt{n} - 1}{\sqrt{n} + 1} - 1\right| < \varepsilon$, $n > N$.

Thus $\lim_{n\to \infty}\frac{\sqrt{n} - 1}{\sqrt{n} + 1} = 1$.

(d) $\lim_{n\to \infty}\frac{n + 1}{n\sqrt{n}}$

If $\varepsilon > 0$ is given, then choosing $N > \frac{1}{\varepsilon^2}$ we have $\frac{1}{\sqrt{n}} < \varepsilon$, $n > N$.

Hence

$$\lim _ {n \to \infty} \frac {1}{\sqrt {n}} = 0.$$

It also true that

$$\lim _ {n \to \infty} \frac {1}{n \sqrt {n}} = 0.$$

Thus,

$$\lim _ {n \to \infty} \frac {n + 1}{n \sqrt {n}} = 0.$$

Answer: (a) 2; (b) 0; (c) 1; (d) 0.

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