Limits and Continuity

Limits

Limits ask the question of what happens to a function $f(x)$ as we get arbitrarily close to a value of $x$. They are denoted as $\lim_{x\rightarrow c}f(x)=L$, where $L, c$ are numbers. In typical cases, we can simply plug in $x=c$ into $f(x)$ to get $L$. To prove that $L$ and $\lim_{x\rightarrow c}f(x)$ are equal, we use the $\delta-\varepsilon$ proof:

\[\begin{matrix} \text{If }\lim_{x\rightarrow c}f(x)=L\text{, then for any }\varepsilon>0\text{, there exists}\\ \text{a }\delta>0\text{ such that }|x-c|<\delta\text{ implies }|f(x)-L|<\varepsilon. \end{matrix}\]

For most functions, the substitution rule typically works: we can simply plug in $x=c$ into $f(x)$ to get $L=f(c)$. However, there are three cases where this wouldn’t work:

  1. We have a vertical asymptote for a rational function. An example is the function $f(x)=1/x$. The limit doesn’t exist in this case.
  2. We have a jump discontinuity that looks like a `step function.’ An example is the function $f(x)=0$ for $x\leq 0$ and $f(x)=1$ for $x>0$. The limit doesn’t exist in this case.
  3. We have infinite oscillations. An example is the function $f(x)=\sin(1/x)$. The limit doesn’t exist in this case.

In cases 1 and 2 above, the reason why the limit doesn’t exist is because the left and right limits are not equal. That is, $\lim_{x\rightarrow c^-} f(x)\neq \lim_{x\rightarrow c^+} f(x)$. This brings up another property of limits:

\[\text{A limit exists if }\lim_{x\rightarrow c^-} f(x)= \lim_{x\rightarrow c^+} f(x)=\lim_{x\rightarrow c} f(x).\]

Limits satisfy pretty straightforward properties:

\[\begin{matrix} \lim_{x\rightarrow c}\left[\alpha f(x)+\beta g(x)\right]=\alpha\lim_{x\rightarrow c}f(x)+\beta\lim_{x\rightarrow c}g(x), \\ \lim_{x\rightarrow c}f(x)g(x)=\left[\lim_{x\rightarrow c}f(x)\right]\left[\lim_{x\rightarrow c}g(x)\right]\\ \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow c}f(x)}{\lim_{x\rightarrow c}g(x)}\text{ as long as }\lim_{x\rightarrow c}g(x)\neq 0 \end{matrix}\]

Limits at infinity are really similar to regular limits, and are related to the idea of horizontal asymptotes. All the same rules apply. Here is an example of a typical way to approach evaluating a limit at infinity:

\[\small\lim_{x\rightarrow \pm\infty} \frac{3}{x-2}=\lim_{x\rightarrow \pm\infty}\frac{3}{x-2}\frac{1/x}{1/x}=\lim_{x\rightarrow \pm\infty}\frac{\frac{3}{x}}{1-\frac{2}{x}}=\frac{0}{1-0}=0\]

Multiplying by the special $(1/x)/(1/x)=1$ term is always determined by the highest power of $x$ in the denominator. We should also talk about the sandwich theorem, a useful technique to evaluate certain limits. Lets consider an example of trying to find the limit of $\lim_{x\to0}x^2\sin(1/x)$. Note that $-1\leq \sin z\leq 1$ for any $z$, meaning that $-x^2\leq x^2\sin x\leq x^2$ for any $x$. $\lim_{x\to0}-x^2=0$ and $\lim_{x\to0}x^2=0$, meaning that $\lim_{x\to0}-x^2=\lim_{x\to0}x^2=0$. Therefore, we can conclude using the sandwich theorem that $\lim_{x\to0}x^2\sin(1/x)=0$. The sandwich theorem also applies for limits at infinity.

Continuity

There’s also the idea of continuity. Functions are continuous if you can trace them without ever lifting your pencil. The formal definition of continuity is that a function $f(x)$ is continuous at $x=c$ if

\[\lim_{x\rightarrow c} f(x)=f(c)\]

For example, all polynomial functions are continuous. There are four types of discontinuities:

  1. Removable discontinuity. You can shift the position of a finite set of points to make the function continuous.
  2. Nonremovable discontinuity: infinite. Think vertical asymptotes of a rational function.
  3. Nonremovable discontinuity: jump. Think of a step function (for example, $f(x)=0$ for $x\leq 0$ and $f(x)=1$ for $x>0$).
  4. Nonremovable discontinuity: oscillatory. Think of $\sin(1/x)$ as $x\rightarrow 0$.

To set up future discussions, consider the slope of a secant line between two points $(x_1, f(x_1))$ and $(x_1+h, f(x_1+h))$ for some parameter $h$. The slope $m$ of this secant line is

\[m=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=\frac{f(x_1+h)-f(x_1)}{h}\]

What happens as $h$ goes to zero? (We’ll find out later.)

Exercises

Problem 1

Evaluate the limit $\lim_{t \rightarrow-3} \frac{6+4 t}{t^{2}+1}$. Evaluate the limit $\lim_{x\rightarrow -\infty}\frac{7x^3+2x+1}{9x^3-4x^2+3x+6}$.

Problem 2

Consider the function

\[f(x)=\left\{\begin{array}{ll} 7-4 x & x<1 \\ x^{2}+2 & x \geq 1 \end{array}\right.\]

Answer the following two questions:

  1. Evaluate the limit $\lim_{x\rightarrow -6}f(x)$ (if it exists). Is $f(x)$ continuous at $x=-6$?
  2. Evaluate the limit $\lim_{x\rightarrow 1}f(x)$ (if it exists). Is $f(x)$ continuous at $x=1$?

Problem 3

Consider the function $f(x)=x^2$.

  1. What is $f(x+h)$ for some parameter $h$? What is $f(x+h)-f(x)$? What is the simplified form of $\frac{f(x+h)-f(x)}{h}$ given this function definition?
  2. Given the simplified form of $\frac{f(x+h)-f(x)}{h}$ you found in Part (1) above, what is $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$?