AI Tutor
High School
SAT
SAT Elite 1500
SAT Tutoring
ACT
ACT Elite 33
ACT Tutoring
University
MCAT
MCAT Elite 515
Med-School Admissions
Med School Calculator
MCAT Score Percentiles
Pre-Med Tutoring
Pre-Med Plus
LSAT
LSAT Elite 170
LSAT Self-Paced
LSAT Tutoring
DAT
DAT Programs
DAT Elite
DAT Self-Paced
DAT Tutoring
Log in
Get Started for Free
One-Sided Limits
Related Topics
Wize University Calculus 1 Textbook > Limits
One-Sided Limits
3 Activities
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
?
Answer
I don't know
Check Submission
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
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
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.
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∣