Use the \(\epsilon \)-\(\delta \) definition of a limit to prove that:

\(\lim\limits_{x\rightarrow c}A=A\), where \(A\) and \(c\) are real numbers

\(\lim\limits_{x\rightarrow c}x=c\), where \(c\) is a real number

Solution

(a) \(f(x)=A\) is the constant function whose graph is a horizontal line. Given any \(\epsilon >0\), we must find \(\delta >0\) so that whenever \(0 \lt \vert x-c\vert \lt \delta ,\) then \(\left\vert f(x)-A\right\vert \lt \epsilon \).

Since \(\left\vert A-A\right\vert =0,\) then \(\left\vert f(x)-A\right\vert \lt \epsilon \) no matter what positive number \(\delta \) is used. That is, any choice of \(\delta \) guarantees that whenever \(0 \lt \vert x-c\vert \lt \delta \), then \(\left\vert f(x)-A\right\vert \lt \epsilon \).

(b) \(f( x) =x\) is the identity function. Given any \( \epsilon >0\), we must find \(\delta \) so that whenever \( 0 \lt \vert x-c\vert \lt \delta \), then \(\vert f(x)-c \vert =\vert x-c\vert \lt \epsilon \). The easiest choice is to make \(\delta =\epsilon \). That is, whenever \(0 \lt \vert x-c\vert \lt \delta =\epsilon \), then \(\vert f(x)-c \vert \underset{\underset{\color{#0066A7}{f( x) =x}}{\color{#0066A7}{\uparrow}}}{=} \vert x-c\vert \lt \epsilon \).