4. Limits of Functions

Let \(c\) be a cluster point of \(A\) and let \(\delta_n=\frac{1}{n}\) for \(n\in\N\). Then by definition of a cluster point, there exists \(x_n\in A\), \(x_n\neq c\), such that \(|x_n-c| \lt \delta_n\). Since \(\delta_n\rightarrow 0\) then \(x_n \rightarrow c\). To prove the converse, suppose that \(x_n \rightarrow c\), \(x_n\neq c\) and \(x_n\in A\) for \(n\in\mathbb{N}\). Then by convergence of \((x_n)\) to \(c\), for any \(\delta \gt 0\) there exists \(K\in\mathbb{N}\) such that \(|x_K-c| \lt \delta\). Since \(x=x_K\in A\), this proves that \(c\) is a cluster point of \(A\).

Suppose that \(f(x)\rightarrow L\) and \(f(x)\rightarrow L'\) as \(x\rightarrow c\), and let \(\varepsilon \gt 0\). Then there exists \(\delta \gt 0\) such that \(|f(x)-L| \lt \eps/2\) and \(|f(x)-L'| \lt \eps/2\), for all \(x\in A\) satisfying \(0 \lt |x-c| \lt \delta\). Then if \(0 \lt |x-c| \lt \delta\) then \begin{align*} |L-L'| &\leq |f(x)-L|+|f(x)-L'|\\ & \lt \eps/2+\eps/2\\ & =\eps. \end{align*} Since \(\eps \gt 0\) is arbitrary, Theorem 2.2.7 implies that \(L=L'\).

We begin by analyzing the quantity \(|f(x)-13|\): \begin{align*} |f(x)-13|&= |5x+3-13|\\ &= |5x-10|\\ & = 5 |x-2|. \end{align*} Hence, if \(0 \lt |x-2| \lt \eps/5\) then \begin{align*} |f(x)-13| &= |5x+3-13|\\ &= 5|x-2|\\ & \lt 5 (\eps/5)\\ & = \eps. \end{align*} Thus, given \(\eps \gt 0\) we let \(\delta = \eps/5\) and thus if \(0 \lt |x-c| \lt \delta\) then \(|f(x) - 13| \lt \eps\). Thus, by definition, \(\lim_{x\rightarrow 2}f(x) = 13\).

We have that \begin{align*} |f(x) - \tfrac{1}{2}| &= \left|\frac{x+1}{x^2+3}-\frac{1}{2}\right| \\[2ex] &= \left|\frac{x^2-2x+1}{2(x^2+3)}\right|\\[2ex] &= \frac{|x-1|^2}{2(x^2+3)} \\[2ex] & \lt |x-1|^2. \end{align*} Let \(\eps \gt 0\) be arbitrary and let \(\delta=\sqrt{\eps}\). Then if \(0 \lt |x-1| \lt \delta\) then \(|x-1|^2 \lt \delta^2 = \eps\). Hence, if \(0 \lt |x-1| \lt \delta\) then \begin{align*} |f(x)-\tfrac{1}{2}| &= \left|\frac{x+1}{x^2+3}-\frac{1}{2}\right| \\[2ex] & \lt |x-1|^2\\ & \lt \eps. \end{align*} Thus, by definition, \(\lim_{x\rightarrow 1}f(x) = \tfrac{1}{2}\).

We first note that \[ |f(x)-c^2| = |x^2-c^2| = |x+c| |x-c|. \] By the triangle inequality, \(|x+c| \leq |x|+|c|\) and therefore \begin{align*} |f(x)-c^2| &= |x+c| |x-c|\\[2ex] & \leq (|x|+|c|) |x-c|. \end{align*} We now need to analyze how large \(|x|\) can become when \(x\) is say within \(\overline{\delta} \gt 0\) of \(c\). To be concrete, suppose that \(\overline{\delta} = 1/2\). Hence, if \(0 \lt |x-c| \lt \overline{\delta}\) then \begin{align*} |x| &= |x-c+c|\\ & \leq |x-c| + |c|\\ & \lt \overline{\delta} + |c|\\ & \lt 1 + |c|. \end{align*} Therefore, if \(0 \lt |x-c| \lt \overline{\delta}\) it holds that \begin{align*} |f(x)-c^2| &\leq (|x|+c) |x-c|\\[2ex] & \lt (1 + c) |x - c|. \end{align*} Now suppose that \(\eps \gt 0\) is arbitrary and let \(\delta=\min\{\overline{\delta}, \frac{\eps}{1+|c|}\}\). Then if \(0 \lt |x-c| \lt \delta\) then \(|x| \lt 1+|c|\) and therefore \begin{align*} |f(x)-c^2| &= |x^2-c^2|\\[2ex] &=|x+c||x-c|\\[2ex] & \leq (|x|+|c|)\cdot \delta\\[2ex] & \lt (1+|c|)\cdot \frac{\eps}{1 + |c|}\\[2ex] & = \eps. \end{align*} This proves, by definition, that \(\lim_{x\rightarrow c} x^2 = c^2\) for any \(c\in\real\).

We first note that \(c=-3\) is indeed a cluster point of \(A=\real\backslash\hspace{-0.3em}\{-3\}\). Now, \begin{align*} |f(x)-2| &= \left|\frac{x^2-3x}{x+3} - 2\right|\\[2ex] &=\left|\frac{x^2-5x-6}{x+3}\right|\\[2ex] &= \left|\frac{(x+1)(x-6)}{(x+3)}\right|\\[2ex] &= \frac{|x+1|}{|x+3|} |x-6|. \end{align*} We now obtain a bound for \(\frac{|x+1|}{|x+3|}\) when \(x\) is close to \(6\). Suppose then that \(|x-6| \lt 1\). Then \(5 \lt x \lt 7\) and therefore, \(6 \lt x+1 \lt 8\), which implies that \(|x+1| \lt 8\). Similarly, if \(|x-6| \lt 1\) then \(8 \lt x+3 \lt 10\) and therefore \(8 \lt |x+3|\), which implies that \(\frac{1}{|x+3|} \lt \frac{1}{8}\). Therefore, if \(|x-6| \lt 1\) then \[ \frac{|x+1|}{|x+3|} \lt 8\cdot \frac{1}{8}=1. \] Suppose now that \(\eps \gt 0\) is arbitrary and let \(\delta=\min\{1,\eps\}\). If \(0 \lt |x-6| \lt \delta\) then from our analysis above it follows that \(\frac{|x+1|}{|x+3|} \lt 1\). Therefore, if \(0 \lt |x-6| \lt \delta\) then \begin{align*} |f(x)-2| &= \left|\frac{x^2-3x}{x+3} - 2\right| \\[2ex] &= \frac{|x+1|}{|x+3|} |x-6|\\[2ex] & \lt 1 \cdot \delta\\ & \leq \eps. \end{align*} This proves that \(\lim_{x\rightarrow 6} f(x) = 2\).

If \(x\in\mathbb{Q}\) then \begin{align*} |f(x)| &= |(x-1)\arctan(x)|\\ &= |x-1| |\arctan(x)|\\ &= |x-1| \cdot \tfrac{\pi}{2} \end{align*} and if \(x\notin\mathbb{Q}\) then \begin{align*} |f(x)| &= \left|\frac{3(x-1)}{1+x^2}\right|\\ &= \frac{3|x-1|}{1+x^2}\\ &\leq 3|x-1|. \end{align*} Therefore, for all \(x\in\real\) it holds that \(|f(x)| \leq 3|x-1|\) since \(\pi/2 \lt 3\). Thus, given \(\eps \gt 0\) let \(\delta = \eps/3\) and thus if \(0 \lt |x-1| \lt \delta\) then \begin{align*} |f(x)| &\leq 3 |x-1|\\ & \lt 3\cdot\delta\\ &= 3\cdot \eps/3\\ &= \eps. \end{align*} This proves that \(\lim_{x\rightarrow 1}f(x) = 0\).

Suppose that \(\lim_{x\rightarrow c} f(x) =L\). Let \((x_n)\) be a sequence in \(A\) converging to \(c\), with \(x_n\neq c\) for all \(n\in\N\). We must prove that the sequence \((f(x_n))\) converges to \(L\). To that end, let \(\eps \gt 0\) be arbitrary. Then, by convergence of \(f\) to \(L\) at \(c\), there exists \(\delta \gt 0\) such that if \(0 \lt |x-c| \lt \delta\) then \(|f(x)-L| \lt \eps\). Now, since \((x_n)\rightarrow c\), there exists \(K\in\N\) such that \(|x_n-c| \lt \delta\) for all \(n\geq K\). Therefore, for \(n\geq K\) we have that \(|f(x_n)-L| \lt \eps\). This proves that \(\lim_{n\rightarrow\infty} f(x_n) = L\). To prove the converse, we prove the contrapositive. Hence, we must show that if \(f\) does not converge to \(L\) then there exists a sequence \((x_n)\) in \(A\) (with \(x_n\neq c\)) converging to \(c\) but the sequence \((f(x_n))\) does not converge to \(L\). Assume then that \(f\) does not converge to \(L\). Then, negating the definition of the limit of a function, there exists \(\eps \gt 0\) such for all \(\delta \gt 0\) there exists \(x\in A\) such that \(0 \lt |x-c| \lt \delta\) and \(|f(x)-L|\geq \eps\). Then, let \(\delta_n=\tfrac{1}{n}\) for \(n\in\N\). Then there exists \(x_n\neq c\) such that \(0 \lt |x_n-c| \lt \delta_n\) and \(|f(x_n)-L|\geq \eps\). Since \(\delta_n\rightarrow 0\) then \((x_n)\rightarrow c\) but clearly \(f(x_n)\) does not converge to \(L\). This ends the proof.

Consider \(x_n=\frac{1}{n}\), which clearly converges to \(c=0\) and \(x_n\neq 0\) for all \(n\in\N\). Then \(f(x_n)= n \) which is unbounded and thus does not converge. Thus, by Corollary 4.1.12, \(\lim_{x\rightarrow 0}\frac{1}{x}\) does not exist.

Let \(f(x)=\sin\left(\tfrac{1}{x}\right)\) with domain \(A=\real\backslash\hspace{-0.3em}\{0\}\). Consider the sequence \(x_n=\frac{1}{\pi/2+n\pi}\). It is clear that \((x_n)\rightarrow 0\) and \(x_n\neq 0\) for all \(n\in\N\). Now \((f(x_n))=(-1,1,-1,-1,\ldots)\) and therefore \((f(x_n))\) does not converge. Therefore, \(f\) has no limit at \(c=0\). In fact, for each \(\alpha\in[0,2\pi)\), consider the sequence \(x_n=\frac{1}{\alpha+2n\pi}\). Clearly \((x_n)\rightarrow 0\) and \(x_n\neq 0\) for all \(n\in\N\). Now, \(f(x_n)=\sin(\alpha+2n\pi)=\sin(\alpha)\). Hence, \((f(x_n))\) converges to \(\sin(\alpha)\). This shows that \(f\) oscillates within the interval \([-1,-1]\) as \(x\) approaches \(c=0\).

Consider the sequence \(x_n=\frac{(-1)^n}{n}\). Then \((x_n)\rightarrow 0\) and \(x_n\neq 0\) for all \(n\in\N\). Now, \(y_n=\text{sgn}(x_n) = (-1)^n\) and thus \((y_n)\) does not converge. Therefore, by Corollary 4.1.12, the function \(\text{sgn}\) has no limit at \(c=0\).

Let \(L=\lim_{x\rightarrow c} f(x)\) and let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta \gt 0\) such that \(|f(x)-L| \lt \eps\) for all \(x\in A\) such that \(0 \lt |x-c| \lt \delta\). Therefore, for all \(x\in A\) and \(0 \lt |x-c| \lt \delta\) we have that \begin{align*} |f(x)| &= |f(x)-L+L|\\ & \leq |f(x)-L| + |L|\\ & \lt \eps +|L|. \end{align*} If \(c\in A\) then let \(M=\max\{|f(c)|, \eps + |L|\}\) and if \(c\notin A\) then let \(M=\eps+|L|\). Then \(|f(x)|\leq M\) for all \(x\in A\) such that \(0 \lt |x-c| \lt \delta\), that is, \(f\) is bounded locally at \(c\).

We cannot use the Limit Laws directly since \(\lim_{x\rightarrow 2} (x-2)=0\). Instead, notice that if \(x\neq 2\) then \(\frac{x^2-4}{x-2}=x+2\). Hence, the functions \(f(x)=\frac{x^2-4}{x-2}\) and \(g(x)=x+2\) are equal at every point in \(\real\backslash\hspace{-0.3em}\{0\}\). It is clear that \(\lim_{x\rightarrow 2} g(x)= 4\) and therefore it follows that also \(\lim_{x\rightarrow 2}f(x) = 4\).

We prove the contrapositive. Suppose then that \(L \lt 0\). Let \(\eps \gt 0\) be such that \(L+\eps \lt 0\). Then since \(\lim_{x\rightarrow c} f(x) = 0\), there exists \(\delta \gt 0\) such that if \(0 \lt |x-c| \lt \delta\) then \(f(x) \lt L+\eps \lt 0\). Hence, \(f(x) \lt 0\) for some \(x\in A\). We give another proof using the sequential criterion for limits. To that end, if \(f\) converges to \(L\) at \(c\) then for any sequence \((x_n)\) converging to \(c\), \(x_n\neq 0\), we have that \(f(x_n)\rightarrow L\). Now \(f(x_n)\geq 0\) and therefore \(L\geq 0\) from our results on limits of sequences (Theorem 3.2.7).

We have that \(0\leq f(x)-M_1\) and therefore by Theorem 4.2.8 we have that \(0\leq L-M_1\). Similarly, from \(0\leq M_2 -f(x)\) we deduce that \(0\leq M_2 - L\). From this we conclude that \(M_1\leq L\leq M_2\). An alternative proof: Since \(f\rightarrow L\) at \(c\), for any sequence \((x_n)\rightarrow c\) with \(x_n\neq 0\), we have that \(f(x_n)\rightarrow L\). Clearly, \(M_1\leq f(x_n)\leq M_2\) and therefore \(M_1\leq L\leq M_2\) from our results on limits of sequences (Theorem 3.2.7).

Let \((x_n)\) be a sequence in \(A\) converging to \(c\) with \(x_n\neq c\) for all \(n\in\N\). Then, by the sequential criterion, \[ L=\lim_{n\rightarrow\infty} g(x_n) = \lim_{n\rightarrow\infty} h(x_n). \] By assumption, it holds that \(g(x_n)\leq f(x_n)\leq h(x_n)\) for all \(n\in\N\), and therefore by the Squeeze Theorem for sequences, we have that \(\lim_{n\rightarrow\infty} f(x_n)=L\). This holds for every such sequence and therefore \(\lim_{x\rightarrow c}f(x) = L\).

Choose \(\eps \gt 0\) so that \(L-\eps \gt 0\), take for example \(\eps=L/2\). Then there exists \(\delta \gt 0\) such that \(L-\eps \lt f(x) \lt L+\eps\) for all \(x\in (c-\delta, c+\delta)\), \(x\neq c\), and thus by transitivity it follows that \(0 \lt f(x)\) for all \(x\in (c-\delta, c+\delta)\), \(x\neq c\).