4. Limits

Let \(c\) be a cluster point of \(A\) and let \(\delta_n=\frac{1}{n}\). Then, there exists \(x_n\in A\), \(x_n\neq c\), such that \(|x_n-c| \lt \delta_n\). Since \(\delta_n\rightarrow 0\) we have that \(\lim x_n = c\). Now suppose that \(\lim x_n=c\) and \(x_n\neq c\), and \((x_n)\) is in \(A\). Then for any \(\delta \gt 0\) there exists \(K\) such that \(|x_n-c| \lt \delta\) for all \(n\geq K\). This proves that \(c\) is a cluster point of \(A\).

Suppose that \(f(x)\rightarrow L\) and \(f(x)\rightarrow L'\) as \(x\rightarrow c\). 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 \(0 \lt |x-c| \lt \delta\). 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 \(\eps\) is arbitrary, Theorem 2.2.7 implies that \(L=L'\).

We begin by analyzing the quantity \(|f(x)-L|\): \begin{align*} |5x+3-13| &= |5x-10|\\ & = 5 |x-2|. \end{align*} Hence, if \(0 \lt |x-2| \lt \eps/5\) then \begin{align*} |5x+3-13| &= 5|x-2|\\ & \lt 5 (\eps/5)\\ & = \eps \end{align*}

We have that \begin{align*} \left|\frac{x+1}{x^2+3}-\frac{1}{2}\right| &= \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 \(|x-1|^2 \lt (\sqrt{\eps})^2 = \eps\). Hence, if \(0 \lt |x-1| \lt \delta\) then \[ \left|\frac{x+1}{x^2+3}-\frac{1}{2}\right| \lt |x-1|^2 \lt \eps. \]

We have that \[ |x^2-c^2| = |x+c| |x-c|. \] By the triangle inequality, \(|x+c| \leq |x|+|c|\) and therefore \begin{align*} |x^2-c^2| &= |x+c| |x-c|\\ & \leq (|x|+|c|) |x-c|. \end{align*} We now obtain a bound for \(|x|+|c|\) when \(x\) is say within \(\delta_1 \gt 0\) of \(c\). Hence, suppose that \(0 \lt |x-c| \lt \delta_1\). Then \begin{align*} |x| &= |x-c+c|\\ & \leq |x-c| + |c|\\ & \lt \delta_1 + |c| \end{align*} and therefore \(|x+c|\leq |x| + |c| \lt \delta_1 +2|c|\). Suppose that \(\eps \gt 0\) is arbitrary and let \(\delta=\min\{\delta_1, \eps/(\delta_1+2|c|)\}\). Then if \(0 \lt |x-c| \lt \delta\) then \(|x+c| \lt \delta_1+2|c|\) and therefore \begin{align*} |x^2-c^2| &=|x+c||x-c|\\ & \lt (\delta_1+2|c|) \delta\\ & \leq (\delta_1+2|c|) \eps/(\delta_1+2|c|)\\ & = \eps. \end{align*}

We have that \begin{align*} \left|\frac{x^2-3x}{x+3} - 2\right| &=\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 \(5 \lt x \lt 7\). Therefore, \(6 \lt x+1 \lt 8\), and therefore \(|x+1| \lt 8\). Similarly, if \(|x-6| \lt 1\) then \(8 \lt x+3 \lt 10\) and therefore \(8 \lt |x+3|\) and therefore \(\frac{1}{|x+3|} \lt \frac{1}{8}\). Therefore, if \(|x-6| \lt 1\) then \[ \frac{|x+1|}{|x+3|} \lt 8\cdot \frac{1}{8}=1. \] Suppose that \(\eps \gt 0\) is arbitrary and let \(\delta=\min\{1,\eps\}\). If \(0 \lt |x-6| \lt \delta\) then from above we have that \(\frac{|x+1|}{|x+3|} \lt 1\), and clearly \(|x-6|\leq \eps\). Therefore, if \(0 \lt |x-6| \lt \delta\) then \begin{align*} \left|\frac{x^2-3x}{x+3} - 2\right| &= \frac{|x+1|}{|x+3|} |x-6|\\ & \lt |x-6|\\ & \leq \eps. \end{align*} This ends the proof.

Suppose that \(f(x)\rightarrow L\) as \(x\rightarrow c\). Let \((x_n)\) be a sequence in \(A\) converging to \(c\), with \(x_n\neq c\). Let \(\eps\) be given. Then 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 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 there exists \(\eps \gt 0\) such for all \(\delta \gt 0\) there exists \(x\) such that \(0 \lt |x-c| \lt \delta\) and \(|f(x)-L|\geq \eps\). Let \(\delta_n=\tfrac{1}{n}\). Then there exists \(x_n\neq c\) such that \(0 \lt |x_n-c| \lt \delta_n\) and \(|f(x_n)-L|\geq \eps\). Hence \((x_n)\rightarrow c\) but \(f(x_n)\) does not converge to \(L\).

Consider \(x_n=\frac{1}{n}\), which clearly converges to \(c=0\) and \(x_n\neq 0\). Then \(f(x_n)= n \) which does not converge.

Let \(f(x)=\sin\left(\tfrac{1}{x}\right)\). 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 \(n\). 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 \(n\). 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 \(c=0\).

Let \(x_n=\frac{(-1)^n}{n}\). Then \((x_n)\rightarrow 0\) and \(x_n\neq 0\) for all \(n\). Now, \(y_n=\text{sgn}(x_n) = (-1)^n\). Clearly, \((y_n)\) does not converge and thus \(\text{sgn}\) 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 \(\lim_{x\rightarrow 2} (x-2)=0\). Now, 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 \(g(x)=x+2\) agree at every point in \(\real\backslash\hspace{-0.3em}\{0\}\). It is clear that \(\lim_{x\rightarrow 2} g(x)= 4\). Let \(\eps \gt 0\) and let \(\delta \gt 0\) be such that if \(0 \lt |x-2| \lt \delta\) then \(|g(x)-4| \lt \eps\). Then for \(0 \lt |x-2| \lt \delta\) we have that \(|f(x)-4|=|g(x)-4| \lt \eps\). Hence \(\lim_{x\rightarrow 2}f(x) = 4\).

We prove the contrapositive. Suppose that \(L \lt 0\). Let \(\eps \gt 0\) be such that \(L+\eps \lt 0\). 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 \(x\in A\). An alternative proof: 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.

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 \(M_1\leq L\leq M_2\).

Let \((x_n)\) be a sequence in \(A\) converging to \(c\), \(x_n\neq c\). Then \(L=\lim g(x_n) = \lim h(x_n)\). Clearly \(g(x_n)\leq f(x_n)\leq h(x_n)\), and therefore by the Squeeze Theorem for sequences we have that \(\lim f(x_n)=L\). This holds for every such sequence and therefore \(f\rightarrow L\) at \(c\).

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 \(x\in (c-\delta, c+\delta)\), \(x\neq c\).