Download original notes
PAGE 1

2.3 Computing Limits

last time: find limits by graph or making table of values

today: by analyzing the functions algebraically

first, easy ones:

If \( f(x) = 3 \) then clearly,

\[ \lim_{x \to 5} f(x) = \lim_{x \to 5} 3 = 3 \]

because 3 is not affected by \( x \)

generalize: (a number "constant")

\[ \lim_{x \to a} C = C \]

another easy one:

if \( f(x) = x \) then

\[ \lim_{x \to 5} f(x) = \lim_{x \to 5} x = 5 \]

generalize:

\[ \lim_{x \to a} x = a \]
PAGE 2

with those and some basic rules ("laws") we can calculate almost any limit

for example,

\[ \begin{aligned} \lim_{x \to 5} (3x) &= \lim_{x \to 5} [(3)(x)] \\ &= \left( \lim_{x \to 5} 3 \right) \left( \lim_{x \to 5} x \right) = 3 \cdot 5 = 15 \quad \text{"Product Law"} \end{aligned} \]

generalize: \( \lim_{x \to a} cx = c \cdot a \)

likewise for quotients:

\[ \lim_{x \to 5} \left( \frac{3}{x} \right) = \frac{\lim_{x \to 5} 3}{\lim_{x \to 5} x} = \frac{3}{5} \quad \text{"Quotient Law"} \]

powers and roots:

\[ \lim_{x \to 5} x^3 = \left( \lim_{x \to 5} x \right)^3 = 5^3 = 125 \quad \text{"Power Law"} \]
\[ \lim_{x \to 5} \sqrt{x} = \sqrt{\left( \lim_{x \to 5} x \right)} = \sqrt{5} \quad \text{"Root Law"} \]
PAGE 3

Limit Properties: Additions and Subtractions

Additions / Subtractions work as expected, too.

Example

\[ \lim_{x \to -2} (x^3 + 3x - 5) \]\[ = \lim_{x \to -2} x^3 + \lim_{x \to -2} 3x - \lim_{x \to -2} 5 \]\[ = \left( \lim_{x \to -2} x \right)^3 + \left( \lim_{x \to -2} 3 \right) \left( \lim_{x \to -2} x \right) - \lim_{x \to -2} 5 \]\[ = (-2)^3 + (3)(-2) - 5 = -19 \]

Looks like we simply plugged in \( x = -2 \) (the number \( x \) is approaching) even though for limits we don't really care if \( x \) can be that number.

We can make the function as close to \( -19 \) as we want by making \( x \) sufficiently close to \( -2 \).

\( x^3 + 3x - 5 \) is a polynomial function \( \rightarrow \) domain is all reals.

So, for polynomials, because there are no breaks, the limit is easy: simply plug in the number \( x \) is approaching.

PAGE 4

Limits of Polynomials

Polynomial has no breaks. As \( x \to a \), \( f(x) \to f(a) \).

In fact, \( \lim_{x \to a} f(x) = f(a) \) for polynomials.

Coordinate graph showing a smooth curve f(x) with points approaching (a, f(a)) from both sides.

Limits of Rational Functions

For rational functions (polynomial over polynomial), we can do the same if \( a \) in \( x \to a \) is in the domain.

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)} \quad \text{if } a \text{ is in the domain} \]

Example

\[ \lim_{x \to 1} \frac{x^2 + x - 20}{x - 4} \]

\( \frac{x^2 + x - 20}{x - 4} \) is a rational function whose domain is \( (-\infty, 4) \cup (4, \infty) \).

\( x \to 1 \) and \( 1 \) is in the domain, so:

\[ \lim_{x \to 1} \frac{x^2 + x - 20}{x - 4} = \frac{1^2 + 1 - 20}{1 - 4} = \frac{-18}{-3} = 6 \]
PAGE 5

Introduction to Indeterminate Forms

What about \[ \lim_{x \to 4} \frac{x^2 + x - 20}{x - 4} ? \]

Clearly, \( \frac{x^2 + x - 20}{x - 4} \) is not defined at \( x = 4 \).

But the limit of \( \frac{x^2 + x - 20}{x - 4} \) may still exist as \( x \to 4 \).

(\( \frac{x^2 + x - 20}{x - 4} \) may still approach some number as \( x \) gets close to 4)

What if we plug in \( x = 4 \)?

\[ \frac{4^2 + 4 - 20}{4 - 4} = \frac{0}{0} = ? \]

"indeterminate form"

\[ \frac{0}{0} \text{ means } \frac{\text{very small number}}{\text{very small number}} \]

We handle this by doing some factoring and canceling.

PAGE 6

Solving Limits by Factoring

\[ \lim_{x \to 4} \frac{x^2 + x - 20}{x - 4} = \lim_{x \to 4} \frac{(x + 5)(x - 4)}{(x - 4)} \]

This is what causes \( \frac{0}{0} \)

We can cancel out \( x - 4 \) since \( x - 4 \) is never zero.

(Remember, \( x \to 4 \) means \( x \) is close to 4 but NOT equal to 4)

\[ \lim_{x \to 4} \frac{(x + 5)(x - 4)}{(x - 4)} = \lim_{x \to 4} (x + 5) = 4 + 5 = 9 \]

polynomial

Graphical Representation

If \( x \neq 4 \), \( \frac{x^2 + x - 20}{x - 4} = x + 5 \)

At \( x = 4 \), \( \frac{x^2 + x - 20}{x - 4} \) is not defined.

That's why there is a hole.

Graph of the linear function y = x + 5 showing a hole at the point (4, 9) on the Cartesian plane.

It is NOT correct to say \( \frac{x^2 + x - 20}{x - 4} = x + 5 \) because for two functions to be equal, they must be equal at ALL \( x \).

PAGE 7

Another way to handle \(\frac{0}{0}\)

Example

\[ \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} \]

never an acceptable answer to limit question

if \(x = 0\) then \(\frac{\sqrt{4} - 2}{0} = \frac{0}{0} = ?\)

there is nothing to factor / cancel

root: rationalize

\[ = \lim_{x \to 0} \left( \frac{\sqrt{4+x} - 2}{x} \cdot \frac{\sqrt{4+x} + 2}{\sqrt{4+x} + 2} \right) \]
\[ = \lim_{x \to 0} \frac{(\sqrt{4+x})^2 + 2\sqrt{4+x} - 2\sqrt{4+x} - 4}{x(\sqrt{4+x} + 2)} \]
\[ = \lim_{x \to 0} \frac{4+x-4}{x(\sqrt{4+x}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{4+x}+2)} = \lim_{x \to 0} \frac{1}{\sqrt{4+x}+2} = \frac{1}{\sqrt{4}+2} \]
\[ = \frac{1}{4} \]
PAGE 8

Example

\[ \lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x-2} \]

if \(x = 2\) then \(\frac{\frac{1}{2} - \frac{1}{2}}{2-2} = \frac{0}{0} = ?\)

do NOT stop at \(\frac{0}{0}\)

\[ = \lim_{x \to 2} \frac{\frac{2}{2x} - \frac{x}{2x}}{x-2} = \lim_{x \to 2} \frac{\frac{2-x}{2x}}{x-2} \]

\(2-x = -(x-2)\)

\[ = \lim_{x \to 2} \frac{2-x}{2x} \cdot \frac{1}{x-2} = \lim_{x \to 2} \frac{-(x-2)}{2x} \cdot \frac{1}{(x-2)} \]
\[ = \lim_{x \to 2} \frac{-1}{2x} = -\frac{1}{4} \]
PAGE 9

Squeeze / Sandwich Theorem

If \( f(x) \le g(x) \le h(x) \) for some \( x \) near \( x = a \), except possibly at \( x = a \),

then

\[ \lim_{x \to a} f(x) \le \lim_{x \to a} g(x) \le \lim_{x \to a} h(x) \]
Graph showing three functions f(x), g(x), and h(x) where g(x) is squeezed between f(x) and h(x) at point a.