Rational Functions and
Indeterminant Limits
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2020
Rationa Functions; Indeterminant Limits - Exercises
What is a rational function?
Any number that can be written as a fraction is rational. In a similar manner of
definition, any function that can be written as a fraction of functions, is called a
rational function.
If f(x) is a polynomial function then
lim
x→a
f(x) = f (a)
However, sometimes when you take a limit of a rational function you end up with
something like
0
0
which makes no sense. This is an indeterminant form. We can use
different techniques to try and evaluate this kind of limits.
Example
Let’s consider an example. What is the following limit,
lim
x→0
√
x + 1 − 1
x
Solution: This limit is of the
0
0
indeterminant form. Let’s evaluate this limit.
lim
x→0
√
x + 1 − 1
x
√
x + 1 + 1
√
x + 1 + 1
, Rationalize the numerator
= lim
x→0
x + 1 − 1
x(
√
x + 1 − 1)
= lim
x→0
x
x(
√
x + 1 + 1)
= lim
x→0
1
√
x + 1 + 1
=
1
1 + 1
=
1
2
Example
Let’s try another example. Evaluate the following limit,
lim
x→1
x − 1
√
x − 1
1
Rationa Functions; Indeterminant Limits - Exercises
Solution:
lim
x→1
x − 1
√
x − 1
= lim
x→1
x − 1
√
x − 1
√
x + 1
√
x + 1
= lim
x→1
(x − 1)(
√
x + 1)
x − 1
, Rationalize the denominator
= lim
x→1
(
√
x + 1)
= 2
Another type of example of an indeterminate form is the following,
lim
x→0
(x + 8)
1/3
− 2
x
We’ll use a substitution to evaluate the above limit. In particular, let u = (x + 8)
1/3
.
Then, u
3
= x + 8 and u
3
− 8 = x. As x → 0 we have u → 2. So our limit becomes,
lim
u→2
u − 2
u
3
− 8
= lim
u→2
u − 2
(u − 2)(u
2
+ 2u + 4)
= lim
u→2
1
u
2
+ 2u + 4
= lim
u→2
1
4 + 4 + 4
=
1
12
One more example.
lim
x→2
|x − 2|
x − 2
Since we’re dealing with the absolute value function, we need to consider the limit
from he left and right hand sides.
lim
x→2
−
|x − 2|
x − 2
and lim
x→2
+
|x − 2|
x − 2
2
Rationa Functions; Indeterminant Limits - Exercises
Let’s consider the function,
f(x) =
|x − 2|
x − 2
=
x−2
x−2
if x > 2
−
(x−2)
x−2
if x < 2
=
1 if x > 2
−1 if x < 2
Therefore,
lim
x→2
−
|x − 2|
x − 2
= −1 and lim
x→2
+
|x − 2|
x − 2
= 1.
Since,
lim
x→2
−
|x − 2|
x − 2
6= lim
x→2
+
|x − 2|
x − 2
the limit,
lim
x→2
|x − 2|
x − 2
does not exist. Let’s consider a couple of more straight forward examples.
lim
x→−1
x
2
− 5x + 2
2x
3
+ 3x + 1
=
lim
x→−1
(x
2
− 5x + 2)
lim
x→−1
(2x
3
+ 3x + 1)
=
1 + 5 + 2
−2 − 3 + 1
=
8
−4
= −2
The different ways to evaluate indeterminate forms of limits include,
1. direct substitution
2. factoring
3. rationalizing
4. change of variable
5. one-sided limits
Now that we have a formal definition and understanidng of the limit of a function,
we can define continuity of a function at a point.
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