MA005: Calculus I

Practice 1: (a) $3-1 < 4 x-5 < 3+1$ so $7 < 4 x < 9$ and $1.75 < x < 2.25:$ " $x$ within $1 / 4$ unit of $\mathrm{2}$".

(b) $\quad 3-0.08 < 4 \mathrm{x}-5 < 3+0.08$ so $7.92 < 4 \mathrm{x} < 8.08$ and $1.98 < \mathrm{x} < 2.02: "\mathrm{x}$ within $0.02$ units of $\mathrm{2}$".

(c) $3-\mathrm{E} < 4 \mathrm{x}-5 < 3+\mathrm{E}$ so $8_{-} \mathrm{E} < 4 \mathrm{x} < 8+\mathrm{E}$ and $2-\frac{\overline{\mathrm{E}}}{4} < \mathrm{x} < 2+{\mathrm{E}}{4}$: "x within $\mathrm{E} / 4$ units of $\mathrm{2}$".

Practice 2: "within 1 unit of 3": If $2 < \sqrt{x} < 4$, then $4 < x < 16$ which extends from 5 units to the left of 9 to 7 units to right of 9. Using the smaller of these two distances from 9, "If $x$ is within 5 units of 9, then $\sqrt{x}$ is within $\mathrm{1}$ unit of $\mathrm{3}$".

"within $0.2$ units of $3$": If $2.8 < \sqrt{x} < 3.2$, then $7.84 < x < 10.24$ which extends from $1.16$ units to the left of 9 to $1.24$ units to the right of 9. "If $x$ is within $1.16$ units of 9, then $\sqrt{x}$ is wqithin $0.2$ units of $\mathrm{3}$.

Practice 3: See Fig. 22.

Practice 4: See Fig. 23.

Practice 5: Given any $\varepsilon>0$, take $\delta=\varepsilon / 5$

If $x$ is within $\delta=\varepsilon / 5$ of 4, then

$\begin{array}{ll}-\varepsilon / 5 < \mathrm{x}-4 < \varepsilon / 5 & & \text { (subtracting 4) } \\ -\varepsilon < 5 \mathrm{x}-20 < \varepsilon & & \text { (multiplying by 5) } \\ -\varepsilon < (5 \mathrm{x}+3)-23 < \varepsilon & & \text { (rearranging to get the form we want)}\end{array}$

We have shown that "for any $\varepsilon > 0$, there is a $\delta > 0$ (namely $\delta=\varepsilon / 5)$" so that the rest of the definition is satisfied.

Practice 6: This is a much more sophisticated (= harder) problem.

Using "proof by contradiction" as outlined in the solution to Example 6. 

(i) Assume that the limit exists: $\lim _{\mathrm{x} \rightarrow 0} \frac{1}{x}=\mathrm{L}$ for some value for $\mathrm{L}$. Let $\varepsilon=1$. (The definition says "for every ε" so we can pick this value. For this limit, the definition fails for every $\varepsilon > 0.$) Then, since we are assuming that the limit exists, there is a $\delta > 0$ so that if $x$ is within $\delta$ of 0 then $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}$ is within $\varepsilon=1$ of L.

(ii) (See Fig. 24) Let $x_{1}$ be between $\mathrm{0}$ and $0+\delta$ and also require that $x_{1} < \frac{1}{2}$. Then $0 < \mathrm{x}_{1} < \frac{1}{2}$ so $\mathrm{f}\left(\mathrm{x}_{1}\right)=\frac{1}{\mathrm{x}_{1}} > 2$. Since $\mathrm{x}_{1}$ is within $\delta$ of $0, \mathrm{f}\left(\mathrm{x}_{1}\right) > 2$ is within $\varepsilon=1$ of $\mathrm{L}$, so L is greater than $2-\varepsilon=1: \mathbf{1} < \mathbf{L}$.

Let $\mathrm{x}_{2}$ be between 0 and $0-\delta$ and also require that $\mathrm{x}_{2} > -\frac{1}{2}$. Then $0 > \mathrm{x}_{2} > \frac{1}{2}$ so $\mathrm{f}\left(\mathrm{x}_{2}\right)=\frac{1}{\mathrm{x}_{2}} < -2$. Since $\mathrm{x}_{2}$ is within $\delta$ of $0, \mathrm{f}\left(\mathrm{x}_{2}\right) < -2$ is

$\varepsilon=1$ of $\mathrm{L}$, so $\mathrm{L}$ is less than $-2+\varepsilon=-1:-\mathbf{1} > \mathbf{L}$.

(iii) The two inequalities in bold print provide the contradiction we hoping to find, There is no value $\mathrm{L}$ that satisfies

BOTH $1 < \mathbf{L}$ and $\mathbf{L} < -\mathbf{1}$

so we can conclude that our assumption was false and that $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}} \,$ does not have a limit as $\mathrm{x} \rightarrow 0$.