Finding Limits Algebraically (aka finding limits analytically)

Goal: To be able to solve for limits without a graph or table of values by the algebraic methods of (1) direct substitution, (2) factoring, (3) rationalization, and (4) resolving a complex fraction.

Use the graph or numerical method to find the value of the following limits. Remember, your answers should be correct to 3 decimal places.

limx→23x + 5 =

limx→−3

ex =

limx→−3

x2 − 35x + 9 =

limx→8log2 x =

limx→23x + 5 =

limx→−3

ex =

limx→−3

x2 − 35x + 9 =

limx→8log2 x =

= 11

= 19.8

= e−3 = .050

= 3

limx→23x + 5 =

limx→−3

ex =

limx→−3

x2 − 35x + 9 =

limx→8log2 x =

= 11

= 19.8

= e−3 = .050

= 3

Notice!!

3(3) + 5 = 11

(−3)2 − 35(−3) + 9 = 19.8

e(−3) = e−3 ≈ .050

log2 (8) = 3

The limits equal the function values for the x-value that is being approached.

When can you use the function value to find the limit?

1. When the function is continuous [not broken} at the x-value you are approaching.

We shall call these “trivial limits.”

Let’s look at a few more examples of trivial limits……..

Important: For limits without breaks into the graph (called continuous functions), find the limit by simply plugging in the x-value into the formula!

All lines have no breaks in them, so just plug 3 into the equation.

When can you use the function value to find the limit?

1. When the function is continuous [not broken} at the x-value you are approaching.

We shall call these “trivial limits.”

2. When the graph has a “hole” at the x-value you are approaching

You check for a hole algebraically (not looking at a graph or a table of values) by DIRECT SUBSTITUTION.

An attempt at using direct substitution to find a "hole"

limx→2

x + 2x2 − 4

=(2) + 2(2)2 − 4

=40

Direct substitution is simply plugging the value in the limit.

40

indicates there is a vertical asymptote at x = 2 (Remember, from Precalculus?)

2. When the graph has a “hole” at the x-value you are approaching

An attempt at using direct substitution to find a "hole"

limx→2

x + 2x2 − 4

=(2) + 2(2)2 − 4

=40

limx→2

x + 2x2 − 4

= Does Not Exist

Let’s try direct substitution on another limit, looking for a “hole.”

limx→3

x − 3x2 − 9

=(3) − 3(3)2 − 9

=00

00

indicates there is a HOLE at x = 3

limx→3

x − 3x2 − 9

=x − 3

(x − 3)(x + 3)= lim

x→3

1x + 3

Why this step?

Let’s expand on that first idea about the equivalent functions….

Because, x − 3x2 − 9

and 1x + 3

are equivalent functions except

at x = 3.

Since they are equivalent functions, we will use 1x + 3

for the limit

because limx→3

1x + 3

is a trivial limit.

limx→3

1x + 3

=1

(3) + 3=16

limx→3

x − 3x2 − 9

=x − 3

(x − 3)(x + 3)= lim

x→3

1x + 3

=16

So, in full…..

Theorem on functions that agree at all but one point (page 56)

Let c be a real number and f(x) = g(x) for all x not equal to c. If the limit of g(x) as x approaches c exists, then the limit of f(x) exists and….

We applied the theorem in the last problem by finding and equivalent function by factoring. Let’s apply this theorem so more….

The two equivalent equations differ only at x = 2, where the first one DNE and the second one equals 1/4

And…

If a function is not “broken” at the x-value we are approaching, then we can just plug that x-value into the equation.

Since f(x) and g(x) agree everywhere but at x = 1, they have the same limits every where. The interesting limit is as x approaches 1.

x3 −1x −1

=(x −1)(x2 + x +1)

x −1= x2 + x +1, but x ≠ 1

lim x→1 f (x) = limx→1 g(x) = (1)2 + (1) +1 = 3

Since x can’t be 3 (break there), let’s find an equivalent unbroken function

Numerical check:

Solving limits algebraically (or analytically)……….. Remember: if you aren’t sure the function is continuous at the x-value you are approaching, try direct substitution and see what you get …..

This is an indeterminant form (0/0), letting us knowthere is a hole at x = -3 and that we need to rewrite the limit by finding an equivalent form.

The most common way of finding the equivalent form is factoring.

Solving limits algebraically (or analytically)………..

Algebraic solution

Graphical check of solution

Graph Check

limx→0

1x= Does Not Exist

Graph Check

This was a trivial limit.

limx→−4

2xx + 4

= Does Not Exist

Factoring provides an equivalent form. Rationalization is another way to find an equivalent form.

We could solve this graphically or numerically (using a table) but rationalization is needed to do this analytically. This is not an intuitive process. Just multiply the top and bottom of the fraction by the conjugate of the radical.

Factoring provides an equivalent form. Rationalization is another way to find an equivalent form.

Now, plug in 0 and we don’t get an indeterminant form, so we have a limit

Checks on the answer…..

Graph check Numeric check

You do ……

Graph provides only an approximate check with an irrational number

When can you use the function value to find the limit?

1. When the function is continuous [not broken} at the x-value you are approaching.

We shall call these “trivial limits.”

2. When the graph has a “hole” at the x-value you are approaching

Direct substitution yields 0/0 then……

Factoring

Rationalization

Simplifying a complex fraction

One more method…..

limx→0

1(x +1)

−1

x=

limx→0

1((0) +1)

−1

(0)=1−10

=00

1(x +1)

−1

(x)=

1(x +1)

− (x +1)(x +1)

x

A complex fraction, but we start the same way….with direct substitution

Let’s find an equivalent form by making this complex fraction a simple one.

limx→0

1(x +1)

−1

x=

1(x +1)

−1

(x)=

1(x +1)

− (x +1)(x +1)

x

1(x +1)

− (x +1)(x +1)

x=

1− x −1x +1x

=

−xx +1x

−xx +1x

=−1x +1

lim x→0−1x +1

=−1

(0) +1= −1

limx→0

1(x +1)

−1

(x)= −1