# Calculus: Limits

## Two-Sided Limits and Limit Laws

If the left-hand limit and a right-hand limit of a function both exist for a particular value and are the same, then the function is said to have a two-sided limit at that value.  Formally:

$$\lim\limits_{x \rightarrow c} f(x) = L \text{ if and only if } \lim\limits_{x\rightarrow c^-} f(x) = \lim\limits_{x \rightarrow c^+} f(x) = L.$$

The phrase "if and only if" means that the implication runs both ways. If a function has a two-sided limit at a value $c$, then the limits from both the left and right at $c$ exist and are equal. Likewise, if a function has limits from both left and right at $c$ and the two are equal, it has a two-sided limit at $c$. The important thing to remember is that we have specifically defined two-sided limits in terms of one-sided limits.

As a cultural note, since this type of limit is of greater interest than one-sided limits, it is often referred to as the limit, without the "two-sided" modifier in front. Regardless, the specific type of limit should be clear from the context.

What about calculating limits directly? The Limit Laws allow us to calculate limits algebraically rather than graphically or numerically. These can be proven using more rigorous mathematics, although such proofs are beyond the scope of this section on calculus. However, the focus in calculus is to acquire an understanding of how to use these rules and apply them in the service of solving problems. The laws are as follows:

1. $\lim\limits_{x \rightarrow c} \left(f(x) + g(x)\right) = \lim\limits_{x \rightarrow c} f(x) + \lim\limits_{x \rightarrow c} g(x)$
2. $\lim\limits_{x \rightarrow c} \left(f(x) - g(x)\right) = \lim\limits_{x \rightarrow c} f(x) - \lim\limits_{x \rightarrow c} g(x)$
3. $\lim\limits_{x \rightarrow c} af(x) = a \lim\limits_{x \rightarrow c} f(x)$
4. $\lim\limits_{x \rightarrow c} f(x) g(x) = \lim\limits_{x \rightarrow c}f(x) \cdot \lim\limits_{x \rightarrow c} g(x)$
5. $\lim\limits_{x \rightarrow c} \dfrac{f(x)}{g(x)} = \dfrac{ \lim\limits_{x \rightarrow c}f(x)}{\lim\limits_{x \rightarrow c} g(x) }$ where $\lim\limits_{x \rightarrow c} g(x) \neq 0$.
6. $\lim\limits_{x \rightarrow c} a = a$
7. $\lim\limits_{x \rightarrow c} x = c$
8. $\lim\limits_{x \rightarrow c} x^{1/n} = c^{1/n}$ where $n \in \mathbb{Z}^+$ and when $n$ is even, $c$ is assumed to be positive.

From the above laws we can derive the following ones:

1. $\lim\limits_{x \rightarrow c} \left(f(x)\right)^n = \left(\lim\limits_{x \rightarrow c} f(x) \right)^n$ where $n \in \mathbb{Z}^+$. This follows from repeated application of #4 above.
2. $\lim\limits_{x \rightarrow c} f(x)^{1/n} = \left(\lim\limits_{x \rightarrow c} f(x)\right)^{1/n}$.

There is an additional law that contains a hint of genius:

1. Let $f(x)$ be of the form $f(x) = \dfrac{g(x)}{g(x)}h(x)$ such that $g(c)=0$. Then $\lim\limits_{x \rightarrow c} f(x) = \lim\limits_{x \rightarrow c} \dfrac{g(x)}{h(x)}$.

The above rule follows from the fact that it does not matter what the value of f(x) is at $c$, or even if it is defined. Therefore we can safely remove the $\dfrac{g(x)}{g(x)}$ term, since it is equal to 1. Note that $g(x)$ could equal $0$ in other locations, but so long as it is not $0$ in a region arbitrarily close to $c$, this isn't an issue. Pathological functions, including the "gotcha" where $g(x)=0$, are not covered by the above rule. Think polynomials, trig functions, and so on.

In the course of evaluating limits, you may find yourself having shown that the limit is equal to $\dfrac{0}{0}$ or $\dfrac{\pm\infty}{\pm\infty}$. These are indeterminate forms, and are a sign that you've messed up and will need to use a different strategy.

Note: If you're interested in learning a more rigorous approach to limits, try picking up a book on real analysis. Advanced calculus books will also have more rigorous exposition of limits. The "$\epsilon-\delta$" definition is what you'll want to look for; proofs of the Limit Laws will be close by. Alternatively, check back at Mathmatique whenever real analysis gets added (although that might take a while).

## Problems

1. Evaluate: $\lim\limits_{x \rightarrow 4} x^2 - 3x$

$\lim\limits_{x \rightarrow 4} x^2 - 3x = 4^2 - 3(4) \\ \lim\limits_{x \rightarrow 4} x^2 - 3x = 16 - 12 \\ \lim\limits_{x \rightarrow 4} x^2 - 3x = 4$

In general, polynomials can be solved by direct substitution. First, break up $x^a$ into $\underbrace{x \cdot x \cdot \ldots \cdot x}_{a}$, then apply rule 4, then rule 6. However, this isn't necessary - just the luxury of enjoy direct substitution.

2. Evaluate: $\lim\limits_{x \rightarrow -h} \dfrac{x+h}{x^2 -2hx + h^2}$

$\lim\limits_{x \rightarrow -h} \dfrac{x+h}{x^2 -2hx + h^2} = \dfrac{-h+h}{(-h)^2 - 2h(-h) + h^2} \\ \lim\limits_{x \rightarrow -h} \dfrac{x+h}{x^2 -2hx + h^2} = \dfrac{0}{h^2 + 2h^2 + h^2} \\ \lim\limits_{x \rightarrow -h} \dfrac{x+h}{x^2 -2hx + h^2} = \dfrac{0}{4h^2} \\ \lim\limits_{x \rightarrow -h} \dfrac{x+h}{x^2 -2hx + h^2} = 0 \\$

3. Evaluate: $\lim\limits_{x \rightarrow 1} \dfrac{x-1}{x^2-1}$

Observe that if we simply try direct substitution, we'll end up with the indeterminate form of $\dfrac{0}{0}$. This means we'll have to be a bit more clever:

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

Note that we can cancel out the $(x-1)$ term by using rule 11 (by letting $g(x)=x-1$).

$\lim\limits_{x \rightarrow 1} \dfrac{x-1}{x^2-1} = \lim\limits_{x \rightarrow 1} \dfrac{1}{x+1} \\ \lim\limits_{x \rightarrow 1} \dfrac{x-1}{x^2-1} = \dfrac{1}{1+1} \\ \lim\limits_{x \rightarrow 1} \dfrac{x-1}{x^2-1} = \dfrac{1}{2} \\$

4. Evaluate: $\lim\limits_{x \rightarrow 3} \dfrac{x^2-5x+6}{x^2-9}$

$\lim\limits_{x \rightarrow 3} \dfrac{x^2-5x+6}{x^2-9} = \lim\limits_{x \rightarrow 3} \dfrac{(x-2)(x-3)}{(x+3)(x-3)} \\ \lim\limits_{x \rightarrow 3} \dfrac{x^2-5x+6}{x^2-9} = \lim\limits_{x \rightarrow 3} \dfrac{x-2}{x+3} \\ \lim\limits_{x \rightarrow 3} \dfrac{x^2-5x+6}{x^2-9} = \dfrac{3-2}{3+3} \\ \lim\limits_{x \rightarrow 3} \dfrac{x^2-5x+6}{x^2-9} = \dfrac{1}{6} \\$

5. Evaluate: $\lim\limits_{x \rightarrow 0} \dfrac{x^2}{x}$

$\lim\limits_{x \rightarrow 0} \dfrac{x^2}{x} = \lim\limits_{x \rightarrow 0} \dfrac{x \cdot x}{x} \\ \lim\limits_{x \rightarrow 0} \dfrac{x^2}{x} = \lim\limits_{x \rightarrow 0} \dfrac{x}{x}\cdot x \\ \lim\limits_{x \rightarrow 0} \dfrac{x^2}{x} = \lim\limits_{x \rightarrow 0} x \\ \lim\limits_{x \rightarrow 0} \dfrac{x^2}{x} = 0 \\$

6. Evaluate: $\lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}}$

$\lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}} = \lim\limits_{x \rightarrow 4} \dfrac{\sqrt{x-4}\sqrt{x-4}}{\sqrt{x-4}} \\ \lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}} = \lim\limits_{x \rightarrow 4} \dfrac{\sqrt{x-4}}{\sqrt{x-4}} \sqrt{x-4} \\ \lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}} = \lim\limits_{x \rightarrow 4} \sqrt{x-4} \\ \lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}} = \sqrt{4-4} \\ \lim\limits_{x \rightarrow 4} \dfrac{x-4}{\sqrt{x-4}} = 0 \\$

7. Evaluate: $\lim\limits_{x \rightarrow 4} 17$

$\lim\limits_{x \rightarrow 4} 17 = 17$

8. Evaluate $\lim\limits_{x \rightarrow 0} \dfrac{1}{x}$

$\lim\limits_{x \rightarrow 0} \dfrac{1}{x} = \dfrac{1}{0} \\ \lim\limits_{x \rightarrow 0} \dfrac{1}{x} = \infty \\$

The limit does not exist.

9. Let the Heaviside step function be denoted by $H(x) = \left\{ \begin{array}[ll] \phantom{}0 & x < 0 \\ 0.5 & x = 0 \\ 1 & x > 0\end{array} \right. \\$

Evaluate $\lim\limits_{x \rightarrow 0}H(x)$.

The trick here is to remember the definition of a two-sided limit. A two-sided limit only exists if both the limit from the left and the right exist and are the same. However, note that the two limits are not the same:

$\lim\limits_{x \rightarrow 0^-} H(x) = 0$

$\lim\limits_{x \rightarrow 0^+} H(x) = 1$

Since $\lim\limits_{x \rightarrow 0^-} H(x) \neq \lim\limits_{x \rightarrow 0^+} H(x)$, the limit does not exist.

10. Evaluate: $\lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

$\lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{\frac{x^2}{x^2}\frac{1}{(x+h)^2} - \frac{(x+h)^2}{(x+h)^2}\frac{1}{x^2}}{h} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{\frac{x^2-(x^2+2hx + h^2)}{x^2(x+h)^2}}{h} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{\frac{x^2-x^2-2hx - h^2}{x^2(x+h)^2}}{h} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{\frac{ -2hx - h^2}{x^2(x+h)^2}}{h} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim\limits_{h \rightarrow 0} \dfrac{-2x - h}{x^2(x+h)^2} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \dfrac{-2x - 0}{x^2(x-0)^2} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \dfrac{-2x}{x^2(x)^2} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \dfrac{-2x}{x^4} \\ \lim\limits_{h \rightarrow 0} \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \dfrac{-2}{x^3} \\$