# MA1S11 (Calculus) - Problems involving the Definition of Limits

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.