Solutions

i)

We have that $|3-\cos(x^2)| \le 4$ for all $x$. Hence, let $\epsilon > 0$ and pick $\delta = \frac{\epsilon}{4}$. This gives

$$|x(3-\cos(x^2))| = |x||3-\cos(x^2)| < \frac{\epsilon}{4}\cdot 4 = \epsilon.$$

and we conclude that $l = 0$.

ii)

We posit that $l = 12$. We have that

$$\begin{align*} |x^2+5x-2-12| &= |x^2+5x-14| \\ &= |x^2-2x+7x-14| \\ &= |x(x-2)+7(x-2)| \\ &= |x-2||x+7|. \end{align*}$$

Hence, for $\epsilon > 0$, choose $\delta = \min\left(\frac{\epsilon}{11},2\right)$, and we obtain

$$|x+7| = |x-2+9| \le |x-2|+9 < 2+9 = 11$$

and

$$|x-2||x+7| < \frac{\epsilon}{11}\cdot 11 = \epsilon$$

for $x$ such that $|x-2| < \delta$.

iii)

We posit that $l = 100$. For $\epsilon > 0$, choose $\delta = \min\left(\frac{\epsilon}{200},\frac{1}{2}\right)$. Hence, by the reverse triangle inequality we have that

$$\begin{align*} 1-|x| &= |x-1| < \frac{1}{2} \\ &\implies \frac{1}{2} < |x| \\ &\implies \frac{1}{|x|} < 2, \end{align*}$$

and

$$\left|\frac{100}{x} - 100\right| = 100\frac{1}{|x|}|x-1| < 100\cdot \frac{\epsilon}{200}\cdot 2 = \epsilon.$$

iv)

We posit that $l = a^4$. Notice that

$$|x^4-a^4| = |x^2-a^2||x^2+a^2| = |x-a||x+a||x^2+a^2|.$$

For $\epsilon > 0$, let $\delta = \min\left(\frac{\epsilon}{(3|a|)(9a^2)}, |a|\right)$. We have that

$$|x|-|a| \le |x-a| < \delta \implies |x|+|a| < \delta+2|a|,$$

so that

$$|x+a| \le |x|+|a| < \delta+2|a| \implies |x+a| < |a|+2|a| = 3|a|.$$

We also gather that

$$|x^2+a^2| = x^2+a^2 \le x^2+2x|a|+a^2 = (|x|+|a|)^2 < 9a^2.$$

Hence,

$$|x^4-a^4| = |x-a||x+a||x^2+a^2| < \frac{\epsilon}{(3|a|)(9a^2)}\cdot 3|a|\cdot 9a^2 < \epsilon.$$

v)

We posit that $l = 2$. Notice that

$$\begin{align*} \left|x^4+\frac{1}{x}-2\right| &= \left|(x^4-1)+\left(\frac{1}{x}-1\right)\right| \\ &\le |x^4-1|+\left|\frac{1}{x}-1\right|. \end{align*}$$

By iv), we know that there is a $\delta_1$ such that when $|x-1|<\delta_1$, $|x^4-1|<\frac{\epsilon}{2}$. Likewise, by iii), there exists a $\delta_2$ such that when $|x-1|<\delta_2$, $\left|\frac{100}{x}-100\right|<50\epsilon$, which is equivalent to $\left|\frac{1}{x}-1\right|<\frac{\epsilon}{2}$. Thus, letting $\delta = \min(\delta_1,\delta_2)$, we have that

$$|x-1|<\delta \implies \left|x^4+\frac{1}{x}-2\right| \le |x^4-1|+\left|\frac{1}{x}-1\right| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$

vi)

We posit that $l = 0$. Notice that

$$\begin{align*} 0 \le \sin^2 x \le 1 &\implies -1 \le -\sin^2 x < 0 \\ &\implies 1 \le 2-\sin^2 x < 2 \\ &\implies 1 \le |2-\sin^2 x| \\ &\implies \frac{1}{|2-\sin^2 x|} \le 1. \end{align*}$$

Let $\epsilon > 0$ and $\delta = \epsilon$. Thus,

$$|x| < \epsilon \implies \frac{|x|}{|2-\sin^2 x|} = \left|\frac{x}{2-\sin^2 x}\right| < \epsilon\cdot 1 = \epsilon.$$

vii)

We posit that $l = 0$. Let $\epsilon > 0$ and $\delta = \epsilon^2$. Thus,

$$|x| < \epsilon^2 \implies \sqrt{|x|} = |\sqrt{|x|}| < \epsilon.$$

viii)

We posit that $l = 1$. Let $\epsilon > 0$ and $\delta = \min\left(\frac{3\epsilon}{5},\frac{5}{9}\right)$. We have that

$$\begin{align*} |x-1| \le \frac{5}{9} &\implies -\frac{5}{9} < x-1 \le \frac{5}{9} \\ &\implies \frac{4}{9} \le x \\ &\implies \frac{2}{3} \le \sqrt{x} \\ &\implies \frac{5}{3} \le \sqrt{x}+1 \\ &\implies \frac{1}{|\sqrt{x}+1|} \le \frac{3}{5}. \end{align*}$$

Thus

$$|\sqrt{x}-1| = \frac{|x-1|}{|\sqrt{x}+1|} \le \frac{5}{3}\epsilon\cdot \frac{3}{5} = \epsilon.$$