MA005: Calculus I

Practice 1: As $\mathrm{x}$ becomes arbitrarily large, the values of $\mathrm{f}(\mathrm{x})$ approach $3$ and the values of $\mathrm{g}(\mathrm{x})$ approach $0$.

Practice 2: $\lim \limits_{x \rightarrow \infty} \frac{3 x+4}{x-2}=3$ and $\lim \limits_{x \rightarrow \infty} \frac{\cos (5 x)}{2 x+7}=0$.

Practice 3: The completed table is shown in Fig. 12.

$\begin{array}{r|c|c}x & \frac{6 x+7}{3-2 x} & \frac{\sin (3 x)}{x} \\\hline 10 & -3.94117647 & -0.09880311 \\200 & -3.04030227 & 0.00220912 \\5000 & -3.00160048 & 0.00017869 \\20,000 & \begin{array}{c}-3.00040003\end{array} & 0.00004787 \\\downarrow & \downarrow \\& -3 & 0\end{array}$

Fig. 12

Practice 4: If $x > \sqrt{10} \approx 3.162$, then $f(x)=\frac{1}{x^{2}} < 0.1$.
If $x > \sqrt{1000} \approx 31.62$, then $f(x)=\frac{1}{x^{2}} < 0.001$.
If $x > \sqrt{1 / E}$, then $f(x)=\frac{1}{x^{2}} < E$.

Practice 5 :

(a) $\lim \limits_{x \rightarrow 2^{+}} \frac{7}{2-x}=-\infty$.
As $x \rightarrow 2^{+}$ the values $2-x \rightarrow 0$, and $x > 2$ so $2-x < 0: 2-x$ takes small negative values.
Then the values of $\frac{7}{2-x}=\frac{7}{\text { small negative values }}$ are large negative values so we represent the limit as " $-\infty$ ".
(b) $\lim \limits_{x \rightarrow 2^{+}} \frac{3 x}{2 x-4}=+\infty$.
As $x \rightarrow 2^{+}$ the values of $2 x-4 \rightarrow 0$, and $x > 2$ so $2 x-4 > 0: 2 x-4$ takes small positive values. As $x \rightarrow 2^{+}$ the values of $3 x \rightarrow+6$.
Then the values of $\frac{3 x}{2 x-4}=\frac{\text { values near }+6}{\text { small positive values }}$ are large positive values so we represent the limit as " $+\infty$ ".
(c) $\lim \limits_{x \rightarrow 2^{+}} \frac{3 x^{2}-6 x}{x-2}=6$.
As $x \rightarrow 2^{+}$, the values of $3 x^{2}-6 x \rightarrow 0$ and $x-2 \rightarrow 0$ so we need to do more work. The numerator can be factored $3 \mathrm{x}^{2}-6 \mathrm{x}=3 \mathrm{x}(\mathrm{x}-2)$ and then the rational function can be reduced (since $x \rightarrow 2$ we know $x \neq 2$): $\lim \limits_{x \rightarrow 2^{+}} \frac{3 x^{2}-6 x}{x-2}=\lim \limits_{x \rightarrow 2^{+}} \frac{3 x(x-2)}{x-2}=\lim \limits_{x \rightarrow 2^{+}} 3 x=6$

Practice 6 :

(a) $f(x)=\frac{x^{2}+x}{x^{2}+x-2}=\frac{x(x+1)}{(x-1)(x+2)}$.
$f$ has vertical asymptotes at $\mathrm{x}=1$ and $\mathrm{x}=-2$.
(b) $g(x)=\frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}$.
The value $x=1$ is not in the domain of $g$. If $x \neq 1$, then $g(x)=x+1$ $\mathrm{g}$ has a "hole" when $\mathrm{x}=1$ and no vertical asymptotes.

Practice 7: $\mathrm{g}(\mathrm{x})=2 \mathrm{x}-3+\frac{2}{\mathrm{x}+1}$.
$\mathrm{g}$ has a vertical asymptote at $\mathrm{x}=-1$.
$\mathrm{g}$ has no horizontal asymptotes.
$\lim \limits_{x \rightarrow \infty} \frac{2}{x+1}=0$ so $\mathrm{g}$ has the linear asymptote $\mathrm{y}=2 \mathrm{x}-3$.

Practice 8 :

$f(x)=x^{2}+\frac{2 \cdot \sin (x)}{x}$.
$f$ is not defined at $x=0$, so $f$ has a vertical asymptote or a "hole" when $x=0$.
$\lim \limits_{x \rightarrow 0} x^{2}+\frac{2 \cdot \sin (x)}{x}=0+2=2$ so $f$ has a "hole" when $x=0$.
$\lim \limits_{x \rightarrow \infty} \frac{2 \cdot \sin (x)}{x}=0$ so $\mathrm{f}$ has the nonlinear asymptote $\mathrm{y}=\mathrm{x}^{2}$.