# Calculus: Limits

## One-Sided Limits

The study of limits begins with one-sided limits. First, we'll start with an intuitive definition, then move on to a more formal definition, which we will then explain in detail.

A one-sided limit describes the behavior of a function as its independent variable approaches a particular value from one direction. For example, consider a function $f(x)$. If we see that $f(2)=12$, $f(1.5)=11$, $f(1.25)=10.5$, $f(1.1) = 10.2$, and $f(1.01) = 10.004$, we might reasonably conclude that as $x$ gets closer to $1$, $f(x)$ gets closer to $10$. Since the value of $x$ is getting closer to $1$ coming from the positive direction (that is, as it gets smaller), this is called a right-hand limit. A left-hand limit is similar, but comes from the opposite direction. If we see that $f(0)=8$, $f(0.5)=9.5$, $f(0.9)=9.8$, and $f(0.99) = 9.996$, we see that as $x$ gets closer to $1$, $f(x)$ gets closer to 10.

More formally, a function $f(x)$ has a **left-hand limit** at $x=c$ whose value is $L$ if $x$ is less than $c$ and can be made arbitrarily close to $c$, though not necessarily equalling $c$, and $f(x)$ becomes arbitrarily close to $L$. The right-handed limit is denoted as follows:

$$\lim\limits_{x \rightarrow c^+} f(x) = L$$

A function $f(x)$ has a **right-hand limit** at $x=c$ whose value is $L$ if $x$ is greater than $c$ and can be made arbitrarily close to $c$, though not necessarily equalling $c$, and $f(x)$ becomes arbitrarily close to $L$. The right-handed limit is denoted as follows:

$$\lim\limits_{x \rightarrow c^-} f(x) = L$$

There are a few critical things to keep in mind. One is to remember that what happens when $x=c$ is explicitly unimportant. In fact, $f(c)$ need not even be defined! All that matters is what happens as $x$ gets closer to $c$. Additionally, the limiting value $L$ must be a number. While this may seem obvious (we're studying math, duh!), one must remember that infinity is *not* a number. If it turns out that the limit tends towards infinity, we instead say that the limit **does not exist**, and often abbreviate this finding as **d.n.e.**

So, how does one go about discovering limits? There are a number of ways:

**Graphing:**By graphing a function, we can use our trusty eyeballs to determine what value a function might be approaching. Though not very rigorous, this method is nonetheless useful for quickly determining the neighborhood of a limit, or whether it exists at all.**Building a table of values:**As in the above explanations, one can compute values that are ever closer to the point in question. If after reaching a sufficient level of precision, the function seems to be converging on a set value, we can be reasonably confident in the value of the limit.**Using the Limits Laws:**The Limit Laws are a set set of allowed operations one can do algebraically in order to figure out the value of a limit directly. For "nice" functions, such as polynomials, the simplest such approach is direct substitution of $c$ in for $x$. In functions where this results in an indeterminate form, such as $\dfrac{0}{0}$, additional rules allow for more interesting manipulations that can yield the correct answer.

The problems in this first section on limits will focus on graphing and building tables of values. The next section on two-sided limits will cover the Limit Laws.

## Problems

Find the limit by building a table of values: $\lim\limits_{x \rightarrow \pi/2^+} \tan(x)$

$\begin{array}[ll] \phantom{}x & \tan(x) \\ & \\ \dfrac{\pi}{2}+0.1 & -9.97 \\ \dfrac{\pi}{2}+0.01 & -99.997 \\ \dfrac{\pi}{2}+0.001 & -1000 \\ \dfrac{\pi}{2}+0.0001 & -10000 \\ \end{array}$

We can see that as $x$ approaches $\dfrac{\pi}{2}$ from the positive side, $\tan(x)$ approaches $-\infty$. We can reasonably conclude that the limit does not exist. However, we use the following notation to describe what happens as $x$ gets arbitrarily close to $\dfrac{\pi}{2}$: $$\lim\limits_{x \rightarrow \pi/2^+} \tan(x) = -\infty$$Find the limit by graphing: $\lim\limits_{x \rightarrow \pi/2^-} \tan(x)$

By looking at the graph, we can see that $\lim\limits_{x \rightarrow \pi/2^-} \tan(x) = \infty$.

The limit does not exist.Find the limit by building a table of values: $\lim\limits_{x \rightarrow 0^+} \ln(x)$

$\begin{array}[ll] \phantom{}x & \ln(x) \\ & \\ 0.1 & -2.30 \\ 10^{-5} & -11.51 \\ 10^{-10} & -23.03 \\ 10^{-100} & -230.26 \\ 10^{-1000} & -2302.59 \\ \end{array}$

We can see that as $x$ approaches $0$ from the positive side, $\ln(x)$ approaches $-\infty$. We can reasonably conclude that the limit does not exist. However, we use the following notation to describe what happens as $x$ gets arbitrarily close to $0$: $$\lim\limits_{x \rightarrow 0^+} \ln(x) = -\infty$$Let the Heaviside step function be denoted by $H(x) = \left\{ \begin{array}[ll] \phantom{}0 & x < 0 \\ 0.5 & x = 0 \\ 1 & x > 0\end{array} \right. \\$

Find the limit: $\lim\limits_{x \rightarrow 0^+} H(x)$As $x$ approaches $0$ from the positive side, the value of $H(x)$ is $1$.

We can conclude that $\lim\limits_{x \rightarrow 0^+} H(x) = 1$.

The fact that $H(0)=0.5$ does not impact what the limit is.Find the limit by building a table of values: $\lim\limits_{x \rightarrow 7^+} \dfrac{x^2-1}{x-7}$

$\begin{array}[ll] \phantom{}x & \dfrac{x^2-1}{x-7} \\ & \\ 7.1 & 494.1 \\ 7.01 & 4814.01 \\ 7.001 & 48014 \\ 7.0001 & 480014 \\ \end{array}$

We can see that as $x$ approaches $7$ from the positive side, $\dfrac{x^2-1}{x-7}$ approaches $\infty$. We can reasonably conclude that the limit does not exist. However, we use the following notation to describe what happens as $x$ gets arbitrarily close to $7$: $$\lim\limits_{x \rightarrow 7^+} \dfrac{x^2-1}{x-7} = \infty$$Let $f(x) = \left\{ \begin{array}[ll] \phantom{}x^2 + 2 & x <= 0 \\ \ln(x) & x > 0\end{array} \right. \\$

Graph $f(x)$ to find the following:

- $\lim\limits_{x \rightarrow 0^+} \, f(x)$
- $f( 0 )$
- $\lim\limits_{x \rightarrow 0^-} \, f(x)$

- $\lim\limits_{x \rightarrow 0^+} \, f(x) = -\infty$. The limit does not exist.
- $f( 0 ) = 2$
- $\lim\limits_{x \rightarrow 0^-} \, f(x) = 2$

Let $f(x) = \left\{ \begin{array}[ll] \phantom{}x^3 + 2 & x < 0 \\ x^2 + 2& x > 0\end{array} \right. \\$

Graph $f(x)$ and find the following:

- $\lim\limits_{x \rightarrow 0^+} \, f(x)$
- $f( 0 )$
- $\lim\limits_{x \rightarrow 0^-} \, f(x)$

- $\lim\limits_{x \rightarrow 0^+} \, f(x) = 2$
- $0$ is not in the domain of $f$, therefore $f(0)$ is undefined.
- $\lim\limits_{x \rightarrow 0^-} \, f(x) = 2$

Let $f(x) = \left\{ \begin{array}[ll] {} \cos(x) & x < 0 \\ 0 & x = 0 \\ \sin(x + \pi/4) & x > 0 \end{array} \right. \\$

Graph $f(x)$ and find the following:

- $\lim\limits_{x \rightarrow 0^+} \, f(x)$
- $f( 0 )$
- $\lim\limits_{x \rightarrow 0^-} \, f(x)$

- $\lim\limits_{x \rightarrow 0^+} \, f(x) = \dfrac{\sqrt{2}}{2}$
- $f(0) = 0$
- $\lim\limits_{x \rightarrow 0^-} \, f(x) = 1$

Let $f(x) = \dfrac{x^2}{x^3+1}$

Graph $f(x)$ and find the following:

- $\lim\limits_{x \rightarrow -1^+} \, f(x)$
- $\lim\limits_{x \rightarrow -1^-} \, f(x)$

- $\lim\limits_{x \rightarrow -1^+} \, f(x) = \infty$. The limit does not exist.
- $\lim\limits_{x \rightarrow -1^-} \, f(x) = -\infty$. The limit does not exist.

Find the limit by graphing: $\lim\limits_{x \rightarrow -\pi/2^+} \, \ln\left(\sin\left(x+\dfrac{\pi}{2}\right)\right)$

By looking at the graph, we can see that $\lim\limits_{x \rightarrow -\pi/2^+} \ln\left(\sin\left(x+\dfrac{\pi}{2}\right)\right) = -\infty$.

The limit does not exist.