Problem ST1:
In each of the parts of this question, it is asserted that a function \( f(x) \) of a real variable \( x \) defined for values of \( x \) close to but not necessarily equal to a given real number \( s \) tends to a limit \( L \) as \( x \) tends to \( s \). It must therefore be the case that, given any positive real number \( \varepsilon \), there must exist some positive real number \( \delta \) such that \( |f(x) - L| \lt \varepsilon \) for all real numbers \( x \) that satisfy \( 0 \lt |x - s| \lt \delta \). In each expression, give a formula for \( \delta \) in terms of \( \varepsilon \), valid for values of \( \varepsilon \) satisfying \( 0 \lt \varepsilon \lt 1 \), that ensures that the required inequalities are satisfied.
For example, in the case of the limit \( \lim\limits_{x \to 0} 2 \root 3 \of{|x|} = 0 \), it suffices to take \( \delta = {\textstyle\frac{1}{8}} \varepsilon^3 \).
(a) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \), to verify that \( \lim\limits_{x \to 4} (3x + 4) = 16 \).
Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{3}} \varepsilon \).
In fact any value of \( \delta \) satisfying \( 0 \lt \delta \leq {\textstyle\frac{1}{3}} \varepsilon \) is a valid solution for this problem. Indeed if if \( 0 \lt |x - 4| \lt \delta \) then \( 4 - \delta \lt x \lt 4 + \delta \) and hence \( 16 - 3 \delta \lt 3x + 4 \lt 16 + 3 \delta \). Thus if \( \delta \leq {\textstyle\frac{1}{3}} \varepsilon \) then \( 16 - \varepsilon \lt 3x + 4 \lt 16 + \varepsilon \) for all real numbers \( x \) satisfying \( 0 \lt |x - 4| \lt \delta \).
(b) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \), to verify that \( \lim\limits_{x \to 0} 4 x^2 = 0 \).
Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{2}} \sqrt{\varepsilon} \).
In fact any value of \( \delta \) satisfying \( 0 \lt \delta \leq {\textstyle\frac{1}{2}} \sqrt{\varepsilon} \) is a valid solution for this problem. Indeed if if \( 0 \lt |x| \lt \delta \), where \( \delta \) is chosen to satisfy \( 0 \lt \delta \leq {\textstyle\frac{1}{2}} \sqrt{\varepsilon} \), then \( 0 \lt 4 x^2 \lt 4 \delta^2 \leq \varepsilon \), and thus the required inequalities are satisfied.
In view of this we could, for example, take \( \delta = {\textstyle\frac{1}{2}} \varepsilon \) (noting that the value of \( \varepsilon \) has been restricted to the range \( 0 \lt \varepsilon \lt 1 \)).
(c) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \), to verify that \( \displaystyle \lim\limits_{x \to 0} \frac{3 \sqrt{|x|}}{1 + x^2} = 0 \).
Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{9}} \varepsilon^2 \).
In fact any value of \( \delta \) satisfying \( 0 \lt \delta \leq {\textstyle\frac{1}{9}} \varepsilon^2 \) is a valid solution for this problem. Indeed if \( -\delta \lt x \lt \delta \), where \( 0 \lt \delta \leq {\textstyle\frac{1}{9}} \varepsilon^2 \), then \[ 0 \leq \frac{3 \sqrt{|x|}}{1 + x^2} \leq 3 \sqrt{|x|} \lt 3 \sqrt{\delta} \leq \varepsilon.\]
(d) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \), to verify that \( \lim\limits_{x \to 2} f(x) = 2 \), where \[ f(x) = \left\{ \begin{array}{ll} 6 - 2x &\mbox{if $x \leq 2$;} \\ 5x - 8 &\mbox{if $x \gt 2$.} \end{array} \right. \]
Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{5}} \varepsilon \).
Indeed if \( 2 - \frac{1}{2} \varepsilon \lt x \lt 2 \) then \( 2 \leq 6 - 2x \lt 2 + \varepsilon \), and if \( 2 \lt x \lt 2 + \frac{1}{5} \varepsilon \) then \( 2 \leq 5x - 8 \lt 2 + \varepsilon \). We should choose \( \delta \gt 0 \) so that both \( \delta \leq {\textstyle\frac{1}{2}} \varepsilon \) and \( \delta \leq {\textstyle\frac{1}{5}} \varepsilon \). We should therefore choose \( \delta \) so that \( 0 \lt \delta \leq {\textstyle\frac{1}{5}} \varepsilon \).
(e) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \), to verify that \( \lim\limits_{x \to 3} g(x) = 0 \), where \[ g(x) = \left\{ \begin{array}{ll} 100 (x - 3)^2 &\mbox{if $x \leq 3$;} \\ 2 \root 3 \of{x - 3} &\mbox{if $x \gt 3$.} \end{array} \right. \]
Solution The required inequalities are satisfied with \( \delta = \mathrm{minimum}( {\textstyle\frac{1}{10}} \sqrt{\varepsilon}, {\textstyle\frac{1}{8}} \varepsilon^3 ) \).
Alternative Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{10}} \varepsilon^3 \) (where \( \varepsilon \) is restricted to the range \( 0 \lt \varepsilon \lt 1 \)).
Indeed if \( 0 \lt \delta \lt \frac{1}{10} \sqrt{\varepsilon} \), and if \( 3 - \delta \lt x \lt 3 \) then \( 0 \leq 100 (x - 3)^2 \lt \varepsilon \). Also if \( 0 \lt \delta \lt {\textstyle\frac{1}{8}} \varepsilon^3 \), and if \( 3 \lt x \lt 3 + \delta \) then \( 0 \leq {\textstyle\frac{1}{2}} \root 3 \of {x - 3} \lt \varepsilon \). Moreover these values of \( \delta \) are the largest that will work as \(x \) approaches \( 3 - \delta \) and \( 3 + \delta \). We therefore need to ensure that \( \delta \leq \mathrm{minimum}( {\textstyle\frac{1}{10}} \sqrt{\varepsilon}, {\textstyle\frac{1}{8}} \varepsilon^3 ) \). This yields the first solution. To obtain the second solution we use the fact that \( \varepsilon \) is restricted to the range \( 0 \lt \varepsilon \lt 1 \), and therefore \( \varepsilon^3 \lt \sqrt{\varepsilon} \), so that \[ {\textstyle\frac{1}{10}} \varepsilon^3 \leq \mathrm{minimum}( {\textstyle\frac{1}{10}} \sqrt{\varepsilon}, {\textstyle\frac{1}{8}} \varepsilon^3 ) \].
(f) Determine \( \delta \gt 0 \) in terms of \( \varepsilon \), where \( 0 \lt \varepsilon \lt 1 \) to verify that \( \lim\limits_{x \to 1} h(x) = 2 \), where \[ h(x) = \left\{ \begin{array}{ll} 2x &\mbox{if $x \leq 1$;} \\[3pt] \displaystyle\frac{x^2 - 1}{x - 1} &\mbox{if $x \gt 1$.} \end{array} \right. \]
Solution The required inequalities are satisfied with \( \delta = {\textstyle\frac{1}{2}} \varepsilon \).
Indeed if \( x \gt 1 \) then \( h(x) = x + 1 \). It follows that if \( 1 \lt x \lt 1 + \delta \), where \( \delta \leq \varepsilon\), then \[ 2 \leq h(x) \lt 2 + \delta \leq 2 + \varepsilon.\] And if \( 1 - \delta \lt x \lt 1 \), where \( \delta \leq {\textstyle\frac{1}{2}} \varepsilon \), then \[ 2 - \varepsilon \leq 2 - 2 \delta \lt h(x) \leq 2.\] It follows that the required inequalities will be satisfied for a value of \( \delta \) that satisfies \( 0 \lt \delta \leq {\textstyle\frac{1}{2}} \varepsilon \).
Problem ST2:
Let \( f(x) \) be a function of a real variable \( x \) that is defined for all non-zero values of \( x \) that lie sufficiently close to zero. We say that \( f(x) \) increases without limit as \( x \) tends to zero, and write \( f(x) \to +\infty \) as \( x \to 0 \), if, given any positive real number \(K\), there exists some positive real number \( \delta \) such that \( f(x) \gt K \) for all real numbers \( x \) that satisfy \( 0 \lt |x| \lt \delta \).
(a) Show, by finding a value of \(\delta\), dependent on \(K\), that ensures that the necessary inequalities are satisfied, that \( \displaystyle \frac{1}{x^2} \) increases without limit as \( x \) tends to zero.
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Solution. Let some positive real number \(K\) be given. Take \( \displaystyle \delta = \frac{1}{\sqrt{K}} \). If \( x \) is a non-zero real number satisfying \( -\delta \lt x \lt \delta \) then \( 0 \lt x^2 \lt \delta^2 \) and therefore \[ \frac{1}{x^2} \gt \frac{1}{\delta^2} = K. \quad\mbox{Q.E.D.}\]
Alternative Solution. Let some positive real number \(K\) be given. Take \[ \displaystyle \delta = \mathrm{minimum} \left( 1, \frac{1}{K} \right).\] If \( x \) is a non-zero real number satisfying \( -\delta \lt x \lt \delta \) then \( |x| \lt 1 \) and \( \displaystyle |x| \lt \frac{1}{K} \) and therefore \( \displaystyle x^2 = |x|^2 \lt |x| \lt \frac{1}{K}.\) But then \( \displaystyle \frac{1}{x^2} \gt K\). Q.E.D.
(b) Let \( f(x) \) and \( g(x) \) be functions of a real variable \( x \) defined for all non-zero values of \( x \) that lie sufficiently close to zero, and let \( C \) be a positive real number. Suppose that \( f(x) \) increases without limit as \( x \) tends to zero. Prove that \( f(x) - C \) increases without limit as \( x \) tends to zero.
Solution Let some positive real number \(K\) be given. Then there exists some positive number \( \delta \) such that \( f(x) \gt K + C \) for all real numbers \(x\) satisfying \( 0 \lt |x| \lt \delta \). But then \( f(x) - C \gt K \) for all real numbers \(x\) satisfying \( 0 \lt |x| \lt \delta \). Thus \( f(x) - C \to +\infty \) as \( x \to 0 \), as required. Q.E.D.
(c) Let \( f(x) \) and \( g(x) \) be functions of a real variable \( x \) defined for all non-zero values of \( x \) that lie sufficiently close to zero. Suppose that \( f(x) \) increases without limit as \( x \) tends to zero. Suppose also that there exist positive constants \(A\) and \( u \) such that \( g(x) \geq A \) for all non-zero real numbers \(x\) satisfying \( -u \lt x \lt u \). Prove that \( f(x) g(x) \) increases without limit as \( x \) tends to zero.
Solution Let some positive real number \(K\) be given. Then there exists some positive number \( \delta_1 \) such that \( f(x) \gt K / A \) for all real numbers \(x\) satisfying \( 0 \lt |x| \lt \delta_1 \). Let \( \delta \) be the minimum of \( \delta_1 \) and \( u \). If \( x \) is a real number satisfying \( 0 \lt |x| \lt \delta \) then \( f(x) \gt K / A \gt 0 \) and \( g(x) \gt A \gt 0 \), and therefore \( f(x) g(x) \gt K \). Thus \( f(x) g(x) \to +\infty \) as \( x \to 0 \), as required. Q.E.D.