Properties of Limits

There are many rules for computing limits. I'll list the most important ones. There are analogous results for left and right-hand limits; just replace "$\displaystyle \lim_{x\to c}$ " with "$\displaystyle \lim_{x\to c^+}$ " or "$\displaystyle \lim_{x\to c^-}$ ".

I mentioned the first rule earlier:

$$\lim_{x \to c} f(x) = f(c).$$

Remember that a polynomial is a function of the form

$$a_0 + a_1 x + \cdots + a_n x^n,$$

where the a's are numbers. The rule says you can compute the limit of a polynomial as x goes to c by plugging c in for x.


Example.

$$\lim_{x \to 1} (x^3 + 2x + 7) = 1 + 2 + 7 = 10.\quad\halmos$$


$$\lim_{x \to a} \left(f(x) + g(x)\right) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x).$$

This equation --- and the ones like it that follow --- must be interpreted in the right way. The statement is true provided that the two limits on the right side are defined.


Example. If I tell you that

$$\lim_{x \to 3} f(x) = 17,$$

then

$$\lim_{x \to 3} \left(f(x) + x^2\right) = 17 + 9 = 26.\quad\halmos$$


$$\lim_{x \to a} \left(f(x)\cdot g(x)\right) = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right).$$

Again, the statement is true provided that the two limits on the right side are defined.

Using mathematical induction and the rule for products, you can prove:

$$\lim_{x \to a} f(x)^n = \left(\lim_{x \to a} f(x)\right)^n.$$

In this equation, n is a positive integer, and the statement is true provided that $\displaystyle \lim_{x
   \to a} f(x)$ is defined.


Example. If I tell you that

$$\lim_{x \to 5} f(x) = 8,$$

then

$$\lim_{x \to 5} 2x^2\cdot f(x) = \left[\lim_{x \to 5} 2x^2\right] \left[\lim_{x \to 5} f(x)\right] = 50\cdot 8 = 400.\quad\halmos$$


You have to be a little careful with quotients to avoid division by zero.

$$\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \hbox{ if } g(a) \ne 0.$$

If the denominator approaches 0, there are two possibilities.

(a) If the numerator approaches a nonzero number, the limit is undefined.

(b) If the numerator approaches 0, you must do additional work to decide whether the limit is defined (and if it's defined, what its value is).

In any case, the statement is only true if $\displaystyle \lim_{x \to a} f(x)$ and $\displaystyle
   \lim_{x \to a} g(x)$ are defined.


Example. Using the rules for quotients and for polynomials,

$$\lim_{x \to 2} \dfrac{x^2 + 3}{2x^3 + 7x - 10} = \dfrac{2^2 + 3}{2\cdot 2^3 + 7\cdot 2 - 10} = \dfrac{7}{20}. \quad\halmos$$


As the last example shows, you can use the rules for quotients and for polynomials to compute the limit of a rational function. A rational function is a polynomial divided by a polynomial. For example,

$$\dfrac{x^2 + 7}{7x^{50} - \sqrt{2}x^3 + 3}, \quad \dfrac{5}{x^7 - 3}, \quad\hbox{and}\quad \dfrac{14x + 1}{x - 2}$$

are rational functions.


Example.

$$\lim_{x\to 2} \dfrac{x^2 + 3x - 4}{x - 2} \quad\hbox{is undefined}.$$

Reason: The numerator $x^2 + 3x - 4$ approaches 6 (a nonzero number) while the denominator $x - 2$ approaches 0.

On the other hand,

$$\lim_{x\to 2} \dfrac{x^2 - 4}{x^2 + 4x + 4} = \dfrac{0}{16} = 0.$$

There's nothing wrong with having 0 on the top of a fraction.


Example. Compute $\displaystyle \lim_{x\to 3} \dfrac{x^2 - 9}{x^2 - 10x + 21}$ .

In this case, plugging in $x = 3$ gives $\dfrac{0}{0}$ , an indeterminate form. I need to do more work to determine whether the limit is defined. Factor and cancel:

$$\lim_{x\to 3} \dfrac{x^2 - 9}{x^2 - 10x + 21} = \lim_{x\to 3} \dfrac{(x - 3)(x + 3)}{(x - 3)(x - 7)} = \lim_{x\to 3} \dfrac{x + 3}{x - 7} = \dfrac{6}{-4} = -\dfrac{3}{2}.\quad\halmos$$


$$\lim_{x \to a} k\cdot f(x) = k\cdot \lim_{x \to a} f(x),$$

provided that the limit on the right is defined.


Example. If you know that

$$\lim_{x\to 4} f(x) = 17, \quad\hbox{then}\quad \lim_{x\to 4} 3\cdot f(x) = 3\cdot \lim_{x\to 4} f(x) = 3\cdot 17 = 51.\quad\halmos$$


$$\lim_{x \to a} \root n \of {f(x)} = \root n \of {\lim_{x \to a} f(x)},$$

provided that n is an odd positive integer, or n is an even positive integer and $\displaystyle \lim_{x \to
   a} f(x)$ is positive.


Example. Compute $\displaystyle \lim_{x\to -1} \root 3 \of {\dfrac{x^2 - 5x -
   2}{x - 2}}$ .

Since I'm taking an odd root, it doesn't matter whether the function inside the root is approaching a positive or a negative number.

$$\lim_{x\to -1} \root 3 \of {\dfrac{x^2 - 5x - 2}{x - 2}} = \root 3 \of {\left(\lim_{x\to -1} \dfrac{x^2 - 5x - 2}{x - 2}\right)} = \root 3 \of {-\dfrac{4}{3}}.\quad\halmos$$


Example. Compute $\displaystyle \lim_{x\to 4} \sqrt{x^2 - 5x - 1}$ .

As $x \to 4$ , the quantity $x^2 -
   5x - 1$ inside the square root approaches $4^2 - 5\cdot 4 - 1 = -5$ . Therefore, the limit is undefined.


Example. Compute $\displaystyle \lim_{x\to 3} \sqrt{x^2 - 9}$ .

As $x \to 3$ , I have $x^2 - 9 \to
   0$ . You might think that, since $\sqrt{0} = 0$ , the limit should be 0. In fact, the limit is undefined.

To see why, remember that when x approaches 3, it does so from both sides. But what happens if x is less than 3? Suppose $x = 2.9$ . Then

$$x^2 - 9 = (2.9)^2 - 9 = 8.41 - 9 = -0.59, \quad\hbox{and}\quad \sqrt{x^2 - 9} = \sqrt{-0.59} \quad\hbox{is undefined}.$$

In other words, x can't approach 3 from the left (through numbers less than 3) because $\sqrt{x^2 - 9}$ is undefined for $x < 3$ . Hence, the limit is undefined.

Later on, I'll show that if x approaches 3 from the right, then the limit is indeed 0.


The next result is often called the Sandwich Theorem (or the Squeeze Theorem). It is different from the other computational rules in that it produces an answer in an indirect way.

The Sandwich Theorem is an intuitively obvious result about limits. Suppose you have three functions $f(x)$ , $g(x)$ , $h(x)$ , and you're trying to compute the limit of $g(x)$ as x approaches a.

Suppose you know that:

1. $\displaystyle \lim_{x \to a} f(x) =
   L$ and $\lim_{x \to a} h(x) = L$ .

2. $f(x) \le g(x) \le h(x)$ (at least for x's in some interval around a).

$$\hbox{\epsfysize=2in \epsffile{limprop1.eps}}$$

Then

$$\lim_{x \to a} g(x) = L.$$

The result in reasonable because g is "sandwiched" between f and h.


Example. $\displaystyle \lim_{x \to 0} x^2 \sin \dfrac{1}{x}$

As $x \to 0$ , $x^2 \to 0$ , but $\sin
   \dfrac{1}{x}$ oscillates. And at $x = 0$ , $\sin \dfrac{1}{x}$ is undefined. There's no "algebraic" rule which would allow you to compute the limit, but the Sandwich Theorem makes it easy.

$\sin (\hbox{anything})$ always lies between -1 and 1:

$$-1 \le \sin \dfrac{1}{x} \le 1.$$

Multiply through by $x^2$ :

$$-x^2 \le x^2 \sin \dfrac{1}{x} \le x^2.$$

Now

$$\lim_{x \to 0} (-x^2) = 0 \quad\hbox{and}\quad \lim_{x \to 0} x^2 = 0.$$

Hence, by the Sandwich Theorem

$$\lim_{x \to 0} x^2 \sin \dfrac{1}{x} = 0.$$

The picture below shows the graphs of $x^2$ , $-x^2$ , and $x^2 \sin \dfrac{1}{x}$ . Notice that $x^2 \sin \dfrac{1}{x}$ is indeed caught between the two parabolas, which squeeze the wiggly graph as $x \to 0$ :

$$\hbox{\epsfysize=2in \epsffile{limprop2.eps}\quad\halmos}$$


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