# Limits - End Behavior and Asymptotes

## Key Questions

Depends on the approaching number and complexity of function.

#### Explanation:

If the function is simple, functions such as $\sin x$ and $\cos x$ are defined for $\left(- \infty , + \infty\right)$ so it's really not that hard.

However, as x approaches infinity, the limit does not exist, since the function is periodic and could be anywhere between $\left[- 1 , 1\right]$

In more complex functions, such as $\sin \frac{x}{x}$ at $x = 0$ there is a certain theorem that helps, called the squeeze theorem. It helps by knowing the limits of the function (eg sinx is between -1 and 1), transforming the simple function to the complex one and, if the side limits are equal, then they squeeze the answer between their common answer. More examples can be seen here.

For $\sin \frac{x}{x}$ the limit as it approaches 0 is 1 (proof too hard), and as it approaches infinity:

$- 1 \le \sin x \le 1$

$- \frac{1}{x} \le \sin \frac{x}{x} \le \frac{1}{x}$

${\lim}_{x \to \infty} - \frac{1}{x} \le {\lim}_{x \to \infty} \sin \frac{x}{x} \le {\lim}_{x \to \infty} \frac{1}{x}$

$0 \le {\lim}_{x \to \infty} \sin \frac{x}{x} \le 0$

Due to the squeeze theorem ${\lim}_{x \to \infty} \sin \frac{x}{x} = 0$

graph{sinx/x [-14.25, 14.23, -7.11, 7.14]}

• The limit of a polynomial function P(x) as x$\to c$ is P(c). That means limit can be found by evaluting P(x) at x=c.

• One confusing, but fundamental, fact about a mathematical limit of a function $f \left(x\right)$ as $x$ approaches some number $c$ is that the value of $f \left(c\right)$ is technically irrelevant, though often useful (when the function is continuous at $c$).

For example, if $f \left(x\right) = \frac{{x}^{2} + 2 x - 3}{x - 1}$ as in the example above, technically the value $f \left(1\right)$ is undefined. However, ${\lim}_{x \to 1} f \left(x\right) = 4$ because the outputs of $f \left(x\right)$ can be made as close to 4 as we want by taking $x$ sufficiently close to, but not equal to, 1. For instance, if we want the value of $f \left(x\right)$ to get within a distance 0.1 of 4, we can take $x$ to be within a distance 0.1 of 1 (note that, for example, $f \left(0.95\right) = 3.95$ and $f \left(1.05\right) = 4.05$).

Why does this happen for this example? Because we can factor the top to get $f \left(x\right) = \frac{\left(x - 1\right) \left(x + 3\right)}{x - 1}$ and then cancel the $x - 1$ factor to say $f \left(x\right) = x + 3$ when $x$ is NOT equal to 1. So the function $f \left(x\right)$ has a graph that is a straight line with a slope of 1 and a y-intercept of 3, except that the point $\left(1 , 4\right)$ is "missing" from the graph (the graph has a "hole" in it). In other words, $f \left(x\right)$ is not continuous at $x = 1$.

A continuous function, whose graph can be drawn without picking up your pencil, such as $f \left(x\right) = {x}^{2}$, can have its limit evaluated as $x$ approaches any number $c$ just by finding $f \left(c\right)$.

An interesting example involving a trigonometric function to consider is ${\lim}_{x \to 0} \frac{\sin \left(x\right)}{x}$. See if you can find this limit and prove that you are right.