MA005: Calculus I

The same type of questions we considered about a function $\mathrm{f}$ as $\mathrm{x}$ approached a finite number can also $b$ asked about $\mathrm{f}$ as $\mathrm{x}$ "becomes arbitrarily large," "increases without bound," and is eventually larger than any fixed number.

Example 1: What happens to the values of $f(x)=\frac{5 x}{2 x+3}$ (Fig. 1) and $\mathrm{g}(x)=\frac{\sin (7 x+1)}{3 x}$ as $\mathrm{x}$ becomes arbitrarily large, as $\mathrm{x}$ increases without bound?

Solution: One approach is numerical: evaluate $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ for some "large" values of $\mathrm{x}$ and see if there is a pattern to the values of $f(x)$ and $g(x)$. Fig. 1 shows the values of $f(x)$ and $g(x)$ for several large values of $x$. When $x$ is very large, it appears that the values of $f(x)$ are close to $2.5=5 / 2$ and the values of $g(x)$ are close to $0$. In fact, we can guarantee that the values of $f(x)$ are as close to $5 / 2$ as someone wants by taking $\mathrm{x}$ to be "big enough".

$\begin{array}{r|r|r}{x} & \frac{5 x}{2 x+3} & \frac{\sin (7 x+1)}{3 x} \\\hline 10 & 2.17 & 0.03170 \\100 & 2.463 & -0.00137 \\1000 & 2.4962 & 0.00033 \\10,000 & 2.4996 & 0.000001\end{array}$

Fig. 1

The values of $\mathrm{f}(\mathrm{x})=\frac{5 \mathrm{x}}{2 \mathrm{x}+3}$ may or may not ever equal $5 / 2$ (they never do), but if $\mathrm{x}$ is "large," then $\mathrm{f}(\mathrm{x})$ is "close to" $5 / 2$. Similarly, we can guarantee that the values of $\mathrm{g}(\mathrm{x})$ are as close to $0$ as someone wants by taking $\mathrm{x}$ to be "big enough". The graphs of $\mathrm{f}$ and $\mathrm{g}$ are shown in Fig. 2 for "large" values of $x$.

Fig. 2

Practice 1: What happens to the values of $f(x)=\frac{3 x+4}{x-2}$ and $g(x)=\frac{\cos (5 x)}{2 x+7}$ as $x$ becomes arbitrarily large?

The answers for Example 1 can be written as limit statements:
"As $\mathrm{x}$ becomes arbitrarily large, the values of $\frac{5 x}{2 x+3}$ approach $\frac{5}{2}^{\prime \prime}$ can be written "$\lim \limits_{x \rightarrow \infty} \frac{5 x}{2 x+3}=\frac{5}{2}$" and "the values of $\frac{\sin (7 \mathrm{x}+1)}{3 \mathrm{x}}$ approach $0$". can be written " $\lim \limits_{x \rightarrow \infty} \frac{\sin (7 x+1)}{3 x}=0$".

The symbol " $\lim \limits_{x \rightarrow \infty}$ " is read "the limit as $\mathrm{x}$ approaches infinity" and means "the limit as $\mathrm{x}$ becomes, arbitrarily large" or as $\mathrm{x}$ increases without bound. (During this discussion and throughout this book, we do not treat "infinity" or "$\infty$", as a number, but only as a useful notation. "Infinity" is not part of the real number system, and we use the common notation " $\mathrm{x} \rightarrow \infty$ " and the phrase " $\mathrm{x}$ approaches infinity" only to mean that "$x$ becomes arbitrarily large". The notation " $\mathrm{x} \rightarrow-\infty$, " read as " $\mathrm{x}$ approaches negative infinity," means that the values of $-\mathrm{x}$ become arbitrarily large.)

Practice 2: Write your answers to Practice 1 using the limit notation.

The $\lim \limits_{x \rightarrow \infty} \mathrm{f}(\mathrm{x})$ asks about the behavior of $\mathrm{f}(\mathrm{x})$ as the values of $\mathrm{x}$ get larger and larger without any bound, and one way to determine this behavior is to look at the values of $\mathrm{f}(\mathrm{x})$ at some values of $x$ which are "large". If the values of the function get arbitrarily close to a single number as $x$ gets larger and larger, then we will say that number is the limit of the function as $x$ approaches infinity. A definition of the limit as " $x \rightarrow \infty$ " is given at the end of this section.

Practice 3: Fill in the table in Fig. 3 for $f(x)=\frac{6 x+7}{3-2 x}$ and $\mathrm{g}(\mathrm{x})=\frac{\sin (3 \mathrm{x})}{\mathrm{x}}$, and then use those values to estimate $\lim \limits_{x \rightarrow \infty} \mathrm{f}(\mathrm{x})$ and $\lim \limits_{x \rightarrow \infty} \mathrm{g}(\mathrm{x})$.

$\begin{array}{r|r|r}x & \frac{6 x+7}{3-2 x} & \frac{\sin (3 x)}{x} \\\hline 10 & & \\200 & & \\5000 & \\20,000 & &\end{array}$

Fig. 3

Example 2: How large does $x$ need to be to guarantee that $f(x)=\frac{1}{x} < 0.1 ? 0.001 ? < E$ (assume $E > 0$)?

Solution: If $x > 10$, then $\frac{1}{x} < \frac{1}{10}=0.1$ (Fig.4). If $x > 1000$, then $\frac{1}{x} < \frac{1}{1000}=0.001$.

In general, if $E$ is any positive number, then we can guarantee that $|f(x)| < E$ by picking only values of $x > \frac{1}{E} > 0$ : if $x > \frac{1}{E}$, then $\frac{1}{x} < E$.

From this we can conclude that $\lim \limits_{x \rightarrow \infty} \frac{1}{x}=0$.

Fig. 4

Practice 4: How large does $\mathrm{x}$ need to be to guarantee that $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}^{2}} < 0.1 ? 0.001 ? < \mathrm{E}$ (assume $\mathrm{E} > 0$)? Evaluate $\lim \limits_{x \rightarrow \infty} \frac{1}{x^{2}}$

The Main Limit Theorem (Section 1.2) about limits of combinations of functions is true if the limits as " $\mathrm{x} \rightarrow \mathrm{a}$ " are replaced with limits as " $x \rightarrow \infty$ ", but we will not prove those results.

Polynomials occur commonly, and we often need the limit, as $\mathrm{x} \rightarrow \infty$, of ratios of polynomials or functions containing powers of $\mathrm{x}$. In those situations the following technique is often helpful:

The limit of the new denominator is a constant, so the limit of the resulting ratio is easier to determine.

Example 3: Determine $\lim \limits_{x \rightarrow \infty} \frac{7 x^{2}+3 x-4}{3 x^{2}-5}$ and $\lim \limits_{x \rightarrow \infty} \frac{9 x+2}{3 x^{2}-5 x+1}$.

Solutions:

$\begin{aligned}&\lim \limits_{x \rightarrow \infty} \frac{7 x^{2}+3 x-4}{3 x^{2}-5}=\lim \limits_{x \rightarrow \infty} \frac{x^{2}\left(7+3 / x-4 / x^{2}\right)}{x^{2}\left(3-5 / x^{2}\right)} \quad \text { factoring out } x^{2} \\&=\lim \limits_{x \rightarrow \infty} \frac{7+3 / x-4 / x^{2}}{3-5 / x^{2}}=\frac{7}{3} \quad \text { canceling } x^{2} \text { and noting } \frac{3}{x}, \frac{4}{x^{2}}, \frac{5}{x^{2}} \rightarrow 0\end{aligned}$

Similarly,

$\begin{aligned}\lim \limits_{x \rightarrow \infty} \frac{9 x+2}{3 x^{2}-5 x+1} &=\lim \limits_{x \rightarrow \infty} \frac{x^{2}\left(9 / x+2 / x^{2}\right)}{x^{2}\left(3-5 / x+1 / x^{2}\right)} \\&=\lim \limits_{x \rightarrow \infty} \frac{9 / x+2 / x^{2}}{3-5 / x+1 / x^{2}}=\frac{0}{3}=0\end{aligned}$

If we have a difficult limit, as $\mathrm{x} \rightarrow \infty$, it is often useful to algebraically manipulate the function into the form of a ratio and then use the previous technique.

If the values of the function oscillate and do not approach a single number as $\mathrm{x}$ becomes arbitrarily large, then the function does not have a limit as $\mathrm{x}$ approaches infinity: the limit Does Not Exist.

Example 4: Evaluate $\lim \limits_{x \rightarrow \infty} \sin (\mathrm{x})$ and $\lim \limits_{x \rightarrow \infty} \mathrm{x}-[\mathrm{x}]$.

Solution: $f(x)=\sin (x)$ and $g(x)=x-[x]$ do not have limits as $x \rightarrow \infty$. As $x$ grows without bound, the values of $f(x)=\sin (x)$ oscillate between $-1$ and $+1$ (Fig. 5), and these values of $\sin (x)$ do not approach a single number. Similarly, $g(x)=x-[x]$ continues to take on values between $0$ and $1$, and these values are not approaching a single number.

Fig. 5