MA005: Calculus I

Practice 1:
(a) $\lim \limits_{x \rightarrow 0} \frac{1-\cos (5 x)}{3 x}$. The numerator and denominator are both differentiable and both equal $0$ when $x=0$, so we can apply l'Hô pital's Rule:

$\lim \limits_{x \rightarrow 0} \frac{1-\cos (5 x)}{3 x}=\lim \limits_{x \rightarrow 0} \frac{5 \cdot \sin (5 x)}{3} \rightarrow \frac{0}{3}=0$

(b) $\lim \limits_{x \rightarrow 2} \frac{x^{2}+x-6}{x^{2}+2 x-8}$. The numerator and denominator are both differentiable functions and they both equal $0$ when $x=0$, so we can apply l'Hô pital's Rule:

$\lim \limits_{x \rightarrow 2} \frac{x^{2}+x-6}{x^{2}+2 x-8}=\lim \limits_{x \rightarrow 2} \frac{2 x+1}{2 x+2}=\frac{5}{6}$

Practice 2: $\lim \limits_{x \rightarrow \infty} \frac{x^{2}+e^{x}}{x^{3}+8 x}$. The numerator and denominator are both differentiable and both become arbitrarily large as $x$ becomes large, so we can apply l'Hô pital's Rule:

$\lim \limits_{x \rightarrow \infty} \frac{x^{2}+e^{x}}{x^{3}+8 x}=\lim \limits_{x \rightarrow \infty} \frac{2 x+e^{x}}{3 x^{2}+8} \rightarrow$ " $\frac{\infty}{\infty}$ ". Using l'Hô pital's Rule again:

$\lim \limits_{x \rightarrow \infty} \frac{2 x+e^{x}}{3 x^{2}+8}=\lim \limits_{x \rightarrow \infty} \frac{2+e^{x}}{6 x} \rightarrow$ " $\frac{\infty}{\infty}$ " and again:

$\lim \limits_{x \rightarrow \infty} \frac{2+e^{x}}{6 x}=\lim \limits_{x \rightarrow \infty} \frac{e^{x}}{6} \rightarrow \infty$.

Practice 3: Comparing $A$ with $\mathrm{e}^{\mathrm{n}}$ operations to $\mathrm{B}$ with $100 \cdot \ln (\mathrm{n})$ operations. $\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{100 \cdot \ln (n)} \rightarrow$ " $\frac{\infty}{\infty}$ " so use L'Hopital's Rule: $\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{100 \cdot \ln (n)}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{100 / n}=\lim \limits_{n \rightarrow \infty} \frac{n \cdot e^{n}}{100}=\infty$ so $B$ requires fewer operations than $A$.

Comparing $B$ with $100 \cdot \ln (\mathrm{n})$ operations to $\mathrm{C}$ with $\mathrm{n}^{5}$ operations.

$\lim \limits_{n \rightarrow \infty} \frac{100 \cdot \ln (n)}{n^{5}} \rightarrow \frac{\infty}{\infty} \cdot \lim \limits_{n \rightarrow \infty} \frac{100 \cdot \ln (n)}{n^{5}}=\lim \limits_{n \rightarrow \infty} \frac{100 / n}{5 n^{4}}=\lim \limits_{n \rightarrow \infty} \frac{100}{5 n^{5}}=0$

so $B$ requires fewer operations than $C$. $B$ requires the fewest operations of the three algorithms.

Comparing $A$ with $\mathrm{e}^{\mathrm{n}}$ operations to $\mathrm{C}$ with $\mathrm{n}^{5}$ operations. Using l'Hô pital's Rule several times:

$\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{n^{5}}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{5 n^{4}}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{20 n^{3}}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{60 n^{2}}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{120 n}=\lim \limits_{n \rightarrow \infty} \frac{e^{n}}{120}=\infty$

so $A$ requires more operations than $C$. $A$ requires the most operations of the three algorithms.