Show the limit does not exist
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.7 Solution: Part aSuppose the limit exists and equals $a$. Then for $\epsilon=\dfrac{1}{2}$, there exists $N>0$ such…
Proof of Squeeze Lemma/Theorem
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.5 Solution: Part a Let $\epsilon>0$. Since $\lim a_n=s$, there exists $N_1>0$ such that $$ |a_n-s|<\epsilon\Longrightarrow -\epsilon…
If the limit of sequence is zero then so is the limit of sequence multiplied by a bounded one
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.4 Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that $$ |s_n-0|=s_n < \frac{\epsilon}{M+1} $$…
If the limit of sequence is zero then so is the limit of square roots
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.3 Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that $$ |s_n-0|=s_n < \epsilon^2 $$…
Determine the limits and prove your claims
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.2 Solution: Part aThe limit is zero. Let $\epsilon>0$ and let $N=\dfrac{1}{\epsilon}$. Then $n>N$ implies$$\frac{n}{n^2+1}<\frac{n}{n^2}=\frac{1}{n}<\frac{1}{N}=\epsilon,$$ and hence$$\left|\frac{n}{n^2+1}-0\right|=\frac{n}{n^2+1}<\epsilon$$…
Formal proof of limits I
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.1 Solution: Part a Let $\epsilon>0$ and let $N=\dfrac{1}{\epsilon}$. Then $n>N$ implies $$ \frac{1}{n}<\frac{1}{N}=\epsilon, $$ and hence…