Question

Q no. 1)Use definition of the limit of a sequence to establish the following limits
A) lim (n/n^2+1)=0
B) lim(2n/n+1)=2
C) lim(n^2-1/2n^2+3)=1/2
D) show that
I) lim (1/3^n)=0 ii) lim( 2^n/n!)=0 iii) lim((2n)^1/n)=1

Answer 1

Answer to Question #82485, Math / Real Analysis

Question

(1) Use definition of limit of a sequence to establish the following limits

(A) $\lim \frac{n}{n^2 + 1} = 0$

(B) $\lim \frac{2n}{n + 1} = 2$

(C) $\lim \frac{n^2 - 1}{2n^2 + 3} = \frac{1}{2}$

(D) Show that

(i) $\lim \frac{1}{3^n} = 0$ (ii) $\lim \frac{2^n}{n!} = 0$ (iii) $\lim (2n)^{\frac{1}{n}} = 1$

Solution

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

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

Let $\epsilon > 0$ be any number and we wish to choose $N$ so that

$$\frac{n}{n^2 + 1} \leq \frac{n}{n^2} = \frac{1}{n} < \epsilon \quad \text{whenever } n > N$$

Thus, the inequality holds if we choose $N > \frac{1}{\epsilon}$

(B) Let $\epsilon > 0$ be any number and we wish to choose $N$ so that

$$\left| \frac{2n}{n + 1} - \frac{2(n + 1)}{n + 1} \right| = \left| \frac{-1}{n + 1} \right| < \frac{1}{n} < \epsilon \quad \text{whenever } n > N$$

Thus, the inequality holds if we choose $N > \frac{1}{\epsilon}$

(C) Let $\epsilon > 0$ be any number and we wish to choose $N$ so that

$$\left| \frac{3n + 1}{2n + 5} - \frac{3}{2} \right| < \left| \frac{3n + 1}{2n} - \frac{3n}{2n} \right| = \frac{1}{2n} < \epsilon \quad \text{whenever } n > N$$

Thus, the inequality holds if we choose $N > \frac{1}{2\epsilon}$

(D)

(i) Consider the result holds for $n$. Then

$$\frac{1}{3^n} < 1$$

Let $\epsilon > 0$ be any number and we wish to choose $N$ so that

$$\left| \frac {1}{3 ^ {n}} \right| < 1 < \epsilon \text{ whenever } n > N$$

Thus, the inequality holds if we choose $N = 1$

(ii) $\frac{2^n}{n!} \leq 2\left(\frac{2}{3}\right)^{n - 2}$ if $n \geq 3$.

If $n = 3$ then $\frac{8}{6} \leq 2\left(\frac{2}{3}\right)$

Consider the result holds for $n$. Then

$$\frac {2 (2 ^ {n})}{(n + 1) n !} \leq \frac {2}{n + 1} \cdot (2) \left(\frac {2}{3}\right) ^ {n - 2} \leq (2) \left(\frac {2}{3}\right) ^ {n - 1}$$

Let $\epsilon > 0$ be any number and we wish to choose $N$ so that

$$\left| \frac {2 ^ {n}}{n !} \right| < \left| \frac {4}{n} \right| < \frac {4}{\left| \frac {4}{n} \right| + 1} < \epsilon \text{ whenever } n > N$$

Thus, the inequality holds if we choose $N = \left[\frac{4}{\epsilon}\right] + 1$

(iii)

$$2 \leq 2 n ^ {\frac {1}{n}} \leq 2 n$$

$n^{1 / n}$ converges to 1, so the limit $\lim (2n)^{\frac{1}{n}}$ converges to 1.

Hence,

$$\lim (2 n) ^ {\frac {1}{n}} = 1$$

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