(last updated: 1:35:08 PM, October 27, 2020)
Definition 3.3.1 Suppose that the function \(f : D \rightarrow\mathbb R\), with \(D\) a subset of \(\mathbb R\) and \(a\) an accumulation point of the set \(D \cap (a,\infty)=\{x \in D \mid x > a \}\) . Then the function \(f\) has a right-hand limit (limit from the right) as \(x\) approaches, tends to, \(a\) if and only if there exists a real number \(L\) such that for every \(\epsilon > 0\) there exists a positive real number \(\delta > 0\) such that \(|f (x) - L|< \epsilon\) provided that \(0<x - a< \delta\) and \(x \in D\). Whenever this is the case, we write \(\displaystyle\lim_{x\rightarrow a+} f (x) = L\). This limit is often denoted by \(f (a+)\), and we say that \(L = f (a+)\) is the right-hand limit of \(f\) at \(a\).
In a similar manner, we can define \(f(a-)\), the left-hand sided limit of \(f\) at \(a\) by replacing “\(0<x - a< \delta\)” with “\(0<a-x< \delta\)” or “\(-\delta< x-a< 0\),” and replacing “\(D \cap (a,\infty)=\{x \in D \mid x > a \}\)” with “\(D \cap (-\infty,a)=\{x \in D \mid x < a \}\).”
Example: Let \(f(x)\) be a polynomial, then \(\displaystyle\lim_{x\rightarrow a+} f (x) =\lim_{x\rightarrow a-} f (x) = f(a)\).
A polynomial is a bounded function on any finite interval. Note that \(g(x)=f(x)-f(x)\) is also a polynomial that has \(x=a\) as a zero, that is, that \(g(a)=0\). This means that \((x-a)\) is a factor of \(g(x)\) so that \(g(x)=(x-a)\cdot h(x)\) for some polynomial \(h(x)\). Let \(M=\max\{|h(x)| \ \mid \ a-1\leq x\leq a+1\}\). Then for any \(\epsilon>0\), let \(\delta=\min\{1,\frac\epsilon M\}\). So when \(|x-a|<\delta\) we have that \(|f(x)-f(x)|=|x-a| \cdot |h(x)|<\delta\cdot M\leq\epsilon\). \(\square\)
Example: Let
\[ f(x)=\begin{cases} 0 &\quad \text{ if } x< 0,x\in\mathbb Q\\ 1 &\quad \text{ if } x< 0,x\not\in\mathbb Q\\ x &\quad \text{ if } x\geq 0 \end{cases} \]
then \(\displaystyle\lim_{x\rightarrow 0+} f (x) = 0\) but \(\displaystyle\lim_{x\rightarrow 0-} f (x)\) does not exist. To see the former, consider any \(\epsilon>0\) and let \(\delta=\epsilon\). Then \(0<x<\delta\) implies \(0<f(x)<\epsilon\). To see the latter, note that for any \(\delta>0\), the interval \((-\delta,0)\) has both rational and irrational numbers, thus \(f(x)\) essentially oscillates between zero and one infinitely fast. It will never approach a limiting value \(L\).
Definition 3.3.2 Suppose that the function \(f : D \rightarrow \mathbb R\), with \(D\) a subset of \(\mathbb R\) and \(a\) an accumulation point of \(D \cap (a, \infty)\). Then the function \(f\) tends to infinity as \(x\) approaches, tends to, \(a\) from the right if and only if for any given real number \(K > 0\), there exists a positive \(\delta> 0\) such that \(f (x) > K\), provided that \(0 <x - a <\delta\) and \(x \in D\). Whenever this is the case, we write \(\lim_{x\rightarrow a^+}f (x) = +\infty\).
Definition 3.3.4. If the limit from the right or from the left at \(x = a\) of a function \(f\) is infinite, meaning \(\pm\infty\), then the line \(x = a\) is called a vertical asymptote.
Example: Let \(f(x)=\frac{x-2}{(x-5)(x-7)}\) and show that \(\lim_{x\rightarrow5^+}f(x)=-\infty\).
We want \(f(x)<-K\) for \(0<x-5<\delta\). Notice that \(x-5\) is in the denominator (and is always positive), so the function approaches \(\frac{(+3)}{(+0)(-2)}\) which is like \(-\infty\).
Note that if \(x\) is close enough (but above) 5, then \(x-2>0\), \(x-5>0\), and \(x-7<0\) (if \(x<7\)). So we wish to pick an initial \(\delta_1\) that will ensure this. Then we can go about finding bounds on different pieces of the function. Thus any \(\delta_1\) that satisfies \(0<\delta_1<2\) will work. If we let \(\delta_1=3\) for example, then \(5<x<5+\delta_1=8\) gives \(x\) values that make \(x-7\) go from positive to zero and negative. That is problematic… even if only for the reason \(x=7\) is not in the domain!
So let \(\delta_1=1\). Then \(5<x<5+\delta_1=6\) gives \(3<x-2<4\) and \(-1<x-7<-2\).
\[ \diamond \S \diamond \]