Limits and continuity
The limit of a function \(f(x)\) at a point \(c\) is the value that the function approaches as \(x\) approaches \(c\). We denote this:
\[ \lim_{x\to c}f(x). \]
If the limit of the function and the actual value of the function agree, then the function is continuous:
\[ \lim_{x\to c}f(x)=f(c) \] If a function is continuous, that makes it especially convenient to evaluate limits because it means you can just plug-in and evaluate the darn function. But as Figure 1 illustrates, not all functions are continuous. Some functions have breaks and jumps and holes that make them discontinuous.
But not all discontinuities are created equal:
- (upper right) the function has a hole. The limit the function approaches (the hole), and the value of the function (the point) are different;
- (lower right) this function is left continuous. The value of the function and the limit as you approach from the left are in agreement. But there’s a jump, and the limit as you approach from the right is different;
- (lower left) this function is right continuous. The value of the function and the limit as you approach from the right are in agreement. But there’s a jump, and the limit as you approach from the left is different;
- (upper left) this function is just straight-up continuous. The left and the right limits both agree with the function value.
Limit rules
Assume \(\lim_{x\to c}f(x)\) and \(\lim_{x\to c}g(x)\) exist, let \(a,\,b\in\mathbb{R}\) be arbitrary constants, and let \(r>0\). Then:
| rule | formula |
|---|---|
| scaling | \[\lim_{x\to c}[a\cdot f(x)]=a\lim_{x\to c}f(x)\] |
| sum/difference | \[\lim_{x\to c}[f(x)\pm g(x)]=\lim_{x\to c}f(x)\pm\lim_{x\to c}g(x)\] |
| linearity | \[\lim_{x\to c}[a\cdot f(x)\pm b\cdot g(x)]=a\lim_{x\to c}f(x)\pm b\lim_{x\to c}g(x)\] |
| product | \[\lim_{x\to c}f(x)g(x)=\left[\lim_{x\to c}f(x)\right]\left[\lim_{x\to c}g(x)\right]\] |
| quotient | \[\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)},\quad \text{if }\lim_{x\to c} g(x)\neq0\] |
| power | \[\lim_{x\to c}f(x)^r=\left[\lim_{x\to c}f(x)\right]^r,\quad \text{if }c>0,\, \lim_{x\to c} f(x)\geq0\] |
Indeterminate forms and L’Hôpital’s rule
Most of the time, we hope that we can appeal to continuity and evaluate limits by simply “plugging-in.” But sometimes plugging in gives loony stuff:
\[ \begin{aligned} \lim_{x\to 0}\frac{\sin x}{x}=1, &\text{ but }\frac{\sin(0)}{0}=\frac{0}{0};\\ \lim_{x\to \infty}\frac{\ln x}{x}=0, &\text{ but }\frac{\ln\infty}{\infty}=\frac{\infty}{\infty}.\\ \end{aligned} \]
Things like \(0/0\), \(\infty/\infty\), and \(0\times\infty\) are examples of indeterminate forms. If you get one, it doesn’t necessarily mean that the limit does not exist. It just means that the jury is still out; it’s indeterminate. To continue investigating, you can use a tool called L’Hôpital’s rule. Whenever \(\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0\) or \(\pm\infty\), the rule says that
\[ \lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}. \] The hope is that the new function \(f'(x)/g'(x)\) will be well-behaved at \(c\), meaning you can simplify by just plugging in. However, after applying L’Hôpital’s rule once and trying to plug-in, you may still get an indeterminate form. In that case, you’ll have to take another round of derivatives, and maybe another after that, until the clouds finally part and the sun shines upon you.
Consider:
\[ \lim_{x\to\infty}\frac{(\ln x)^2}{x}. \]
\(\lim_{x\to\infty}(\ln x)^2=\infty\) and \(\lim_{x\to\infty}x=\infty\), so let’s take some derivatives:
\[ \begin{aligned} \frac{\text{d}}{\text{d}x}(\ln x)^2&=\frac{2\ln x}{x}\\ \frac{\text{d}}{\text{d}x}x&=1. \end{aligned} \] The ratio of derivatives is thus \(2(\ln x)/x\), but this still gives an indeterminate form because \(\lim_{x\to\infty}2\ln x=\infty\) and \(\lim_{x\to\infty}x=\infty\), so we have to go again:
\[ \begin{aligned} \frac{\text{d}}{\text{d}x}2\ln x&=\frac{2}{x}\\ \frac{\text{d}}{\text{d}x}x&=1. \end{aligned} \] The ratio of these derivatives is just \(2/x\), and that goes to 0 as \(x\to \infty\). In sum,
\[ \lim_{x\to\infty}\frac{(\ln x)^2}{x} \overset{\text{L'H}}{=} \lim_{x\to\infty}\frac{2\ln x}{x} \overset{\text{L'H}}{=} \lim_{x\to\infty}\frac{2}{x} = 0. \]