4. Limits of Functions

Let $$ be a cluster point of $$ and let $$ for $$. Then by definition of a cluster point, there exists $$, $$, such that $$. Since $$ then $$. To prove the converse, suppose that $$, $$ and $$ for $$. Then by convergence of $$ to $$, for any $$ there exists $$ such that $$. Since $$, this proves that $$ is a cluster point of $$.

Suppose that $$ and $$ as $$, and let $$. Then there exists $$ such that $$ and $$, for all $$ satisfying $$. Then if $$ then $$ Since $$ is arbitrary, Theorem 2.2.7 implies that $$.

We begin by analyzing the quantity $$: $$ Hence, if $$ then $$ Thus, given $$ we let $$ and thus if $$ then $$. Thus, by definition, $$.

We have that $$ Let $$ be arbitrary and let $$. Then if $$ then $$. Hence, if $$ then $$ Thus, by definition, $$.

We first note that $$ By the triangle inequality, $$ and therefore $$ We now need to analyze how large $$ can become when $$ is say within $$ of $$. To be concrete, suppose that $$. Hence, if $$ then $$ Therefore, if $$ it holds that $$ Now suppose that $$ is arbitrary and let $$. Then if $$ then $$ and therefore $$ This proves, by definition, that $$ for any $$.

We first note that $$ is indeed a cluster point of $$. Now, $$ We now obtain a bound for $$ when $$ is close to $$. Suppose then that $$. Then $$ and therefore, $$, which implies that $$. Similarly, if $$ then $$ and therefore $$, which implies that $$. Therefore, if $$ then $$ Suppose now that $$ is arbitrary and let $$. If $$ then from our analysis above it follows that $$. Therefore, if $$ then $$ This proves that $$.

If $$ then $$ and if $$ then $$ Therefore, for all $$ it holds that $$ since $$. Thus, given $$ let $$ and thus if $$ then $$ This proves that $$.

Suppose that $$. Let $$ be a sequence in $$ converging to $$, with $$ for all $$. We must prove that the sequence $$ converges to $$. To that end, let $$ be arbitrary. Then, by convergence of $$ to $$ at $$, there exists $$ such that if $$ then $$. Now, since $$, there exists $$ such that $$ for all $$. Therefore, for $$ we have that $$. This proves that $$. To prove the converse, we prove the contrapositive. Hence, we must show that if $$ does not converge to $$ then there exists a sequence $$ in $$ (with $$) converging to $$ but the sequence $$ does not converge to $$. Assume then that $$ does not converge to $$. Then, negating the definition of the limit of a function, there exists $$ such for all $$ there exists $$ such that $$ and $$. Then, let $$ for $$. Then there exists $$ such that $$ and $$. Since $$ then $$ but clearly $$ does not converge to $$. This ends the proof.

Consider $$, which clearly converges to $$ and $$ for all $$. Then $$ which is unbounded and thus does not converge. Thus, by Corollary 4.1.12, $$ does not exist.

Let $$ with domain $$. Consider the sequence $$. It is clear that $$ and $$ for all $$. Now $$ and therefore $$ does not converge. Therefore, $$ has no limit at $$. In fact, for each $$, consider the sequence $$. Clearly $$ and $$ for all $$. Now, $$. Hence, $$ converges to $$. This shows that $$ oscillates within the interval $$ as $$ approaches $$.

Consider the sequence $$. Then $$ and $$ for all $$. Now, $$ and thus $$ does not converge. Therefore, by Corollary 4.1.12, the function $$ has no limit at $$.

Let $$ and let $$ be arbitrary. Then there exists $$ such that $$ for all $$ such that $$. Therefore, for all $$ and $$ we have that $$ If $$ then let $$ and if $$ then let $$. Then $$ for all $$ such that $$, that is, $$ is bounded locally at $$.

We cannot use the Limit Laws directly since $$. Instead, notice that if $$ then $$. Hence, the functions $$ and $$ are equal at every point in $$. It is clear that $$ and therefore it follows that also $$.

We prove the contrapositive. Suppose then that $$. Let $$ be such that $$. Then since $$, there exists $$ such that if $$ then $$. Hence, $$ for some $$. We give another proof using the sequential criterion for limits. To that end, if $$ converges to $$ at $$ then for any sequence $$ converging to $$, $$, we have that $$. Now $$ and therefore $$ from our results on limits of sequences (Theorem 3.2.7).

We have that $$ and therefore by Theorem 4.2.8 we have that $$. Similarly, from $$ we deduce that $$. From this we conclude that $$. An alternative proof: Since $$ at $$, for any sequence $$ with $$, we have that $$. Clearly, $$ and therefore $$ from our results on limits of sequences (Theorem 3.2.7).

Let $$ be a sequence in $$ converging to $$ with $$ for all $$. Then, by the sequential criterion, $$ By assumption, it holds that $$ for all $$, and therefore by the Squeeze Theorem for sequences, we have that $$. This holds for every such sequence and therefore $$.

Choose $$ so that $$, take for example $$. Then there exists $$ such that $$ for all $$, $$, and thus by transitivity it follows that $$ for all $$, $$.