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One-sided limits: Evaluating limits by factoring
Related Topics
Wize University Calculus 1 Textbook > Limits
Computing Limits by Factoring
3 Activities
Wize University Calculus 1 Textbook > Limits
One-Sided Limits
3 Activities
Practice: Evaluating limits
Evaluate
lim
x
→
2
−
x
2
−
3
x
+
2
∣
x
−
2
∣
\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
lim
x
→
2
−
∣
x
−
2
∣
x
2
−
3
x
+
2
Answer
I don't know
Check Submission
More Computing Limits by Factoring Questions:
Limits: Factoring Terms
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
Limits: Factoring Terms
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
Limits: Factoring
Evaluate
lim
x
→
27
x
2
/
3
−
9
x
1
/
3
−
3
\lim\limits_{x\to 27} \frac{x^{2/3}-9}{x^{1/3}-3}
x
→
27
lim
x
1/3
−
3
x
2/3
−
9
Limits: Factoring
Evaluate
lim
x
→
0
1
−
cos
2
x
sin
2
2
x
\lim\limits_{x\to 0} \frac{1-\cos 2x}{\sin^22x}
x
→
0
lim
sin
2
2
x
1
−
cos
2
x
Limits: Factoring Terms
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
Limit Techniques: Factoring
lim
x
→
3
x
−
3
x
3
−
27
\displaystyle \lim_{x\rightarrow 3}\frac{x-3}{x^3-27}
x
→
3
lim
x
3
−
27
x
−
3
Limit Techniques: Factoring
lim
x
→
3
x
−
3
x
3
−
27
\displaystyle \lim_{x\rightarrow 3}\frac{x-3}{x^3-27}
x
→
3
lim
x
3
−
27
x
−
3
Limit Techniques: Factoring
lim
x
→
3
x
−
3
x
3
−
27
\displaystyle \lim_{x\rightarrow 3}\frac{x-3}{x^3-27}
x
→
3
lim
x
3
−
27
x
−
3
One-sided limits: Evaluating limits by factoring
Practice: Evaluating limits
Evaluate
lim
x
→
2
−
x
2
−
3
x
+
2
∣
x
−
2
∣
\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
lim
x
→
2
−
∣
x
−
2
∣
x
2
−
3
x
+
2
The limit
lim
x
→
2
x
2
−
5
x
+
6
x
2
−
4
\displaystyle\lim_{x\to2}\frac{x^2-5x+6}{x^2-4}
x
→
2
lim
x
2
−
4
x
2
−
5
x
+
6
evaluates to:
Find the following limit:
lim
x
→
5
x
2
−
25
x
−
5
\displaystyle\lim_{x\rightarrow5}\frac{x^2-25}{x-5}
x
→
5
lim
x
−
5
x
2
−
25
Limits: Factoring
Evaluate
lim
x
→
0
1
−
cos
2
x
sin
2
2
x
\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos{2x}}{\sin^2{2x}}
x
→
0
lim
sin
2
2
x
1
−
cos
2
x
.
Limit with Fractional Powers
Evaluate
lim
x
→
27
x
2
/
3
−
9
x
1
/
3
−
3
\displaystyle \lim_{x\rightarrow 27}\frac{x^{2/3}-9}{x^{1/3}-3}
x
→
27
lim
x
1/3
−
3
x
2/3
−
9
.
Limits: Cubic
Find the limit
lim
x
→
1
x
3
−
1
x
−
1
\displaystyle\lim\limits_{x\to 1}\frac{x^{3}-1}{x-1}
x
→
1
lim
x
−
1
x
3
−
1
Limits: Factoring Terms
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
Limits: Factoring Terms
lim
x
→
2
x
3
−
5
x
2
+
11
x
−
10
x
−
2
\displaystyle\lim_{x\rightarrow2}\frac{x^3-5x^2+11x-10}{x-2}
x
→
2
lim
x
−
2
x
3
−
5
x
2
+
11
x
−
10
lim
x
→
1
x
3
−
14
x
2
+
7
x
+
6
x
−
1
\displaystyle\lim_{x\rightarrow1}\frac{x^3-14x^2+7x+6}{x-1}
x
→
1
lim
x
−
1
x
3
−
14
x
2
+
7
x
+
6
Limits: Factoring Terms
lim
x
→
1
x
9
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\frac{x^9-1}{x-1}
x
→
1
lim
x
−
1
x
9
−
1
Limits: Factoring Terms
lim
x
→
3
x
4
−
81
x
−
3
\displaystyle\lim_{x\rightarrow3}\frac{x^4-81}{x-3}
x
→
3
lim
x
−
3
x
4
−
81
lim
x
→
3
x
4
−
81
x
−
3
\displaystyle\lim_{x\rightarrow3}\frac{x^4-81}{x-3}
x
→
3
lim
x
−
3
x
4
−
81
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
Practice: Factorable Indeterminate Forms
Cancelling Common Factors: Evaluate
lim
x
→
2
x
2
+
x
−
6
x
−
2
\lim_{x\to 2}\limits\dfrac{x^2 + x - 6}{x - 2}
x
→
2
lim
x
−
2
x
2
+
x
−
6
Limits: Factoring
Evaluate
lim
x
→
27
x
2
/
3
−
9
x
1
/
3
−
3
\lim\limits_{x\to 27} \frac{x^{2/3}-9}{x^{1/3}-3}
x
→
27
lim
x
1/3
−
3
x
2/3
−
9
Limits: Factoring
Evaluate
lim
x
→
0
1
−
cos
2
x
sin
2
2
x
\lim\limits_{x\to 0} \frac{1-\cos 2x}{\sin^22x}
x
→
0
lim
sin
2
2
x
1
−
cos
2
x
Limits: Factoring
Evaluate
lim
x
→
0
1
−
cos
2
x
sin
2
2
x
\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos{2x}}{\sin^2{2x}}
x
→
0
lim
sin
2
2
x
1
−
cos
2
x
.
Limit with Fractional Powers
Evaluate
lim
x
→
27
x
2
/
3
−
9
x
1
/
3
−
3
\displaystyle \lim_{x\rightarrow 27}\frac{x^{2/3}-9}{x^{1/3}-3}
x
→
27
lim
x
1/3
−
3
x
2/3
−
9
.
Limits: Cubic
Find the limit
lim
x
→
1
x
3
−
1
x
−
1
\displaystyle\lim\limits_{x\to 1}\frac{x^{3}-1}{x-1}
x
→
1
lim
x
−
1
x
3
−
1
Find the following limit:
lim
x
→
5
x
2
−
25
x
−
5
\displaystyle\lim_{x\rightarrow5}\frac{x^2-25}{x-5}
x
→
5
lim
x
−
5
x
2
−
25
The limit
lim
x
→
2
x
2
−
5
x
+
6
x
2
−
4
\displaystyle\lim_{x\to2}\frac{x^2-5x+6}{x^2-4}
x
→
2
lim
x
2
−
4
x
2
−
5
x
+
6
evaluates to:
The value of the limit
lim
x
→
1
x
−
1
x
2
−
3
x
+
2
\lim_{x\to1}\frac{\sqrt{x}-1}{x^2-3x+2}
lim
x
→
1
x
2
−
3
x
+
2
x
−
1
is
Limits: Factoring (Common Denominator)
Solve the limit
lim
x
→
5
x
−
x
+
10
x
−
2
25
−
x
2
\displaystyle \lim\limits_{x\to 5}\frac{x-\frac{x+10}{x-2}}{25-x^2}
x
→
5
lim
25
−
x
2
x
−
x
−
2
x
+
10
.
Limits: Factoring
What is the limit
lim
x
→
0
sin
2
x
cos
x
1
−
cos
x
\lim\limits_{x\to0}\frac{\sin^2 x\cos x}{1-\cos x}
x
→
0
lim
1
−
c
o
s
x
s
i
n
2
x
c
o
s
x
?
The limit
lim
x
→
2
x
2
−
5
x
+
6
x
2
−
4
\lim\limits_{x\rightarrow 2}\frac{x^2-5x+6}{x^2-4}
x
→
2
lim
x
2
−
4
x
2
−
5
x
+
6
evaluates to
Limits: Factoring terms
Evaluate the following limit
lim
x
→
4
x
(
x
−
2
)
x
−
6
x
+
8
\displaystyle \lim_{x\rightarrow 4}\frac{x(\sqrt{x}-2)}{x-6\sqrt{x}+8}
x
→
4
lim
x
−
6
x
+
8
x
(
x
−
2
)
Evaluate
lim
x
→
2
−
x
2
−
3
x
+
2
∣
x
−
2
∣
\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
lim
x
→
2
−
∣
x
−
2
∣
x
2
−
3
x
+
2
.
🦊
TRICKY!
Determine the value of
lim
x
→
−
1
x
3
+
1
x
+
1
\displaystyle \lim_{x\rightarrow -1}\frac{x^3+1}{x+1}
x
→
−
1
lim
x
+
1
x
3
+
1
.
🦊
TRICKY!
Determine the value of
lim
x
→
2
x
3
−
8
x
−
2
\displaystyle \lim_{x \to 2} \frac{x^3-8}{x-2}
x
→
2
lim
x
−
2
x
3
−
8
.
Evaluate
lim
x
→
∞
4
x
2
+
1
−
2
x
\displaystyle \lim_{x\rightarrow \infty}\sqrt{4x^2+1}-2x
x
→
∞
lim
4
x
2
+
1
−
2
x
.
Evaluate the limit,
lim
x
→
3
x
2
−
9
x
−
3
\displaystyle\lim_{x\rightarrow3}\frac{x^2-9}{x-3}
x
→
3
lim
x
−
3
x
2
−
9
.
Evaluate the limit,
lim
x
→
2
x
+
7
−
3
x
−
2
\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x+7}-3}{x-2}
x
→
2
lim
x
−
2
x
+
7
−
3
.
Evaluate the limit,
lim
x
→
0
x
2
+
x
−
2
−
x
\displaystyle \lim_{x\rightarrow0}\frac{x}{\sqrt{2+x}-\sqrt{2-x}}
x
→
0
lim
2
+
x
−
2
−
x
x
.
Find the following limit:
lim
x
→
3
x
2
−
9
x
−
3
\displaystyle \lim_{x\rightarrow3}\frac{x^2-9}{\sqrt{x}-\sqrt{3}}
x
→
3
lim
x
−
3
x
2
−
9
.
Limits: Factoring
Evaluate the limit
lim
x
→
8
x
3
−
2
x
2
−
64
\displaystyle \lim\limits_{x\to 8}\frac{\sqrt[3]{x}-2}{x^2-64}
x
→
8
lim
x
2
−
64
3
x
−
2
.
Limits by Factoring Terms
Practice: 0/0 Limit
lim
h
→
0
4
−
(
5
h
+
2
)
2
h
=
\displaystyle\lim_{h\to0}\ \frac{4-\left(5h+2\right)^2}{h}=
h
→
0
lim
h
4
−
(
5
h
+
2
)
2
=
Evaluating Limits: Factoring terms
Evaluate the following limit, or state that it does not exist and why.
lim
x
→
1
x
3
−
2
x
2
−
5
x
+
6
x
2
−
1
\lim_{x \rightarrow 1} \frac{x^3 - 2x^2 - 5x + 6}{x^2 - 1}
x
→
1
lim
x
2
−
1
x
3
−
2
x
2
−
5
x
+
6
Limit Techniques: Factoring
lim
x
→
3
x
−
3
x
3
−
27
\displaystyle \lim_{x\rightarrow 3}\frac{x-3}{x^3-27}
x
→
3
lim
x
3
−
27
x
−
3
lim
x
→
1
x
5
−
1
x
−
1
\displaystyle\lim_{x\rightarrow1}\dfrac{x^5-1}{x-1}
x
→
1
lim
x
−
1
x
5
−
1
More One-Sided Limits Questions:
One-sided Limits
Evaluate the limit
lim
x
→
1
x
2
−
1
∣
x
−
1
∣
\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
x
→
1
lim
∣
x
−
1
∣
x
2
−
1
Practice: One-Sided Limits
Find:
a)
lim
x
→
2
+
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
x
→
2
+
lim
x
−
2
∣
x
−
2∣
b)
lim
x
→
2
−
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
x
→
2
−
lim
x
−
2
∣
x
−
2∣
Practice: One-Sided Limits
Find:
a)
lim
x
→
2
+
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
x
→
2
+
lim
x
−
2
∣
x
−
2∣
b)
lim
x
→
2
−
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
x
→
2
−
lim
x
−
2
∣
x
−
2∣
One-sided Limits
Evaluate the limit
lim
x
→
1
x
2
−
1
∣
x
−
1
∣
\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
x
→
1
lim
∣
x
−
1
∣
x
2
−
1
Limit with Absolute Value
Evaluate
lim
x
→
2
x
−
2
∣
5
x
−
10
∣
\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}
x
→
2
lim
∣5
x
−
10∣
x
−
2
, if it exists.
Limit with Absolute Value
Evaluate
lim
x
→
2
x
−
2
∣
5
x
−
10
∣
\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}
x
→
2
lim
∣5
x
−
10∣
x
−
2
, if it exists.
Limit with Absolute Value
Evaluate
lim
x
→
2
x
−
2
∣
5
x
−
10
∣
\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}
x
→
2
lim
∣5
x
−
10∣
x
−
2
, if it exists.
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits as x approaches -1, 2, and 3.
f
(
x
)
=
{
−
3
x
+
5
if
x
<
−
1
−
10
if
x
=
−
1
9
−
x
2
if
−
1
<
x
≤
2
1
−
∣
2
x
−
3
∣
if
x
>
2
f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
3
x
+
5
−
10
9
−
x
2
1
−
∣2
x
−
3∣
if
x
<
−
1
if
x
=
−
1
if
−
1
<
x
≤
2
if
x
>
2
Practice: One-Sided Limits
Find:
a)
lim
x
→
2
+
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
x
→
2
+
lim
x
−
2
∣
x
−
2∣
b)
lim
x
→
2
−
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
x
→
2
−
lim
x
−
2
∣
x
−
2∣
One-sided limits: Evaluating limits by factoring
Practice: Evaluating limits
Evaluate
lim
x
→
2
−
x
2
−
3
x
+
2
∣
x
−
2
∣
\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
lim
x
→
2
−
∣
x
−
2
∣
x
2
−
3
x
+
2
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, what should the limits be as x approaches -1, 2, and 3?
f
(
x
)
=
{
−
3
x
+
5
if
x
<
−
1
−
10
if
x
=
−
1
9
−
x
2
if
−
1
<
x
≤
2
1
−
∣
2
x
−
3
∣
if
x
>
2
f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
3
x
+
5
−
10
9
−
x
2
1
−
∣2
x
−
3∣
if
x
<
−
1
if
x
=
−
1
if
−
1
<
x
≤
2
if
x
>
2
Practice: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f
(
x
)
=
{
−
3
x
+
5
if
x
<
−
1
−
10
if
x
=
−
1
9
−
x
2
if
−
1
<
x
≤
2
1
−
∣
2
x
−
3
∣
if
x
>
2
f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
3
x
+
5
−
10
9
−
x
2
1
−
∣2
x
−
3∣
if
x
<
−
1
if
x
=
−
1
if
−
1
<
x
≤
2
if
x
>
2
Indeterminate Forms: One-sided limits
Evaluate:
lim
x
→
1
−
∣
x
−
1
∣
x
−
1
\lim_{x \rightarrow 1^-} \frac{|x-1|}{x-1}
x
→
1
−
lim
x
−
1
∣
x
−
1∣
One-Sided Limits
Evaluate
lim
x
→
3
−
∣
x
−
3
∣
x
2
−
3
\lim\limits_{x\to 3^-} \frac{|x-3|}{x^2-3}
x
→
3
−
lim
x
2
−
3
∣
x
−
3∣
Practice: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f
(
x
)
=
{
−
3
x
+
5
if
x
<
−
1
−
10
if
x
=
−
1
9
−
x
2
if
−
1
<
x
≤
2
1
−
∣
2
x
−
3
∣
if
x
>
2
f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
3
x
+
5
−
10
9
−
x
2
1
−
∣2
x
−
3∣
if
x
<
−
1
if
x
=
−
1
if
−
1
<
x
≤
2
if
x
>
2
Practice: Limit of a Function
Example: Limit of a Function
The following is the graph of the function
f
(
x
)
f\left(x\right)
f
(
x
)
Determine the following limits:
Consider the function
f
(
x
)
=
{
1
2
x
+
a
if
x
<
2
8
x
2
+
a
−
1
if
x
≥
2
f\left(x\right)=\begin{cases}\dfrac{1}{2}x+a &\text{if} &x<2\\ \dfrac{8}{x^2+a-1}&\text{if} &x\geq2\\ \end{cases}
f
(
x
)
=
⎩
⎨
⎧
2
1
x
+
a
x
2
+
a
−
1
8
if
if
x
<
2
x
≥
2
. What are the values of a for which the limit
lim
x
→
2
f
(
x
)
\displaystyle\lim_{x\to2}f\left(x\right)
x
→
2
lim
f
(
x
)
exists?
Limits: Practice
Find the following limit:
lim
x
→
2
−
5
(
x
+
1
)
x
−
2
\displaystyle\lim_{x\rightarrow 2}\frac{-5(x+1)}{x-2}
x
→
2
lim
x
−
2
−
5
(
x
+
1
)
Limits: Practice
Find the following limit:
lim
x
→
2
−
5
(
x
+
1
)
x
−
2
\displaystyle\lim_{x\rightarrow 2}\frac{-5(x+1)}{x-2}
x
→
2
lim
x
−
2
−
5
(
x
+
1
)
Evaluate the following limits and the value of the function, where the graphs of f(x) and g(x) are given in the figure. If the limit does not exist, input "DNE".
a)
lim
x
→
1
f
(
x
)
=
\displaystyle\lim_{x\rightarrow1}f(x)=
x
→
1
lim
f
(
x
)
=
b)
lim
x
→
2
−
g
(
x
)
=
\displaystyle\lim_{x\rightarrow2^-}g(x)=
x
→
2
−
lim
g
(
x
)
=
c)
lim
x
→
1
+
f
(
x
)
=
\displaystyle\lim_{x\rightarrow1^+}f(x)=
x
→
1
+
lim
f
(
x
)
=
d)
lim
x
→
2
+
g
(
x
)
=
\displaystyle\lim_{x\rightarrow2^+}g(x)=
x
→
2
+
lim
g
(
x
)
=
Let us call all the functions in the figure below as
y
(
x
)
y\left(x\right)
y
(
x
)
. Evaluate the following limits, if they exist. If the limit does not exist, write "DNE".
A)
lim
x
→
−
1
y
(
x
)
\displaystyle\lim_{x\rightarrow-1}\ y(x)
x
→
−
1
lim
y
(
x
)
B)
lim
x
→
2
y
(
x
)
\displaystyle\lim_{x\rightarrow2}\ y(x)
x
→
2
lim
y
(
x
)
Practice: One-Sided Limits
Find:
a)
lim
x
→
2
+
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
x
→
2
+
lim
x
−
2
∣
x
−
2∣
b)
lim
x
→
2
−
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
x
→
2
−
lim
x
−
2
∣
x
−
2∣
One-sided Limits
Evaluate the limit
lim
x
→
1
x
2
−
1
∣
x
−
1
∣
\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
x
→
1
lim
∣
x
−
1
∣
x
2
−
1
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f
(
x
)
=
{
−
3
x
+
5
if
x
<
−
1
−
10
if
x
=
−
1
9
−
x
2
if
−
1
<
x
≤
2
1
−
∣
2
x
−
3
∣
if
x
>
2
f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
−
3
x
+
5
−
10
9
−
x
2
1
−
∣2
x
−
3∣
if
x
<
−
1
if
x
=
−
1
if
−
1
<
x
≤
2
if
x
>
2
Limit with Absolute Value
Evaluate
lim
x
→
2
x
−
2
∣
5
x
−
10
∣
\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}
x
→
2
lim
∣5
x
−
10∣
x
−
2
, if it exists.
One-Sided Limits
If
lim
x
→
a
−
f
(
x
)
=
2
L
+
3
\lim_{x \rightarrow a^-}f(x) = 2L + 3
lim
x
→
a
−
f
(
x
)
=
2
L
+
3
and
lim
x
→
a
+
f
(
x
)
=
3
L
−
1
\lim_{x \rightarrow a^+}f(x) = 3L - 1
lim
x
→
a
+
f
(
x
)
=
3
L
−
1
and
lim
x
→
a
f
(
x
)
\lim_{x \rightarrow a}f(x)
lim
x
→
a
f
(
x
)
exists with
L
L
L
a real number, what is the value of
L
L
L
?
Practice: One-Sided Limits
Find:
a)
lim
x
→
2
+
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
x
→
2
+
lim
x
−
2
∣
x
−
2∣
b)
lim
x
→
2
−
∣
x
−
2
∣
x
−
2
\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
x
→
2
−
lim
x
−
2
∣
x
−
2∣