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Practice: Direct Substitution
Related Topics
Wize University Calculus 1 Textbook > Limits
One-Sided Limits
3 Activities
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
a. i)
lim
x
→
−
1
−
f
(
x
)
=
\displaystyle\lim_{x\to-1^-}f\left(x\right)=
x
→
−
1
−
lim
f
(
x
)
=
a. ii)
lim
x
→
−
1
+
f
(
x
)
=
\displaystyle\lim_{x\to-1^+}f\left(x\right)=
x
→
−
1
+
lim
f
(
x
)
=
a. iii)
lim
x
→
−
1
f
(
x
)
=
\displaystyle\lim_{x\to-1}f\left(x\right)=
x
→
−
1
lim
f
(
x
)
=
b. i)
lim
x
→
2
−
f
(
x
)
=
\displaystyle\lim_{x\to2^-}f\left(x\right)=
x
→
2
−
lim
f
(
x
)
=
b. ii)
lim
x
→
2
+
f
(
x
)
=
\displaystyle\lim_{x\to2^+}f\left(x\right)=
x
→
2
+
lim
f
(
x
)
=
b. iii)
lim
x
→
2
f
(
x
)
=
\displaystyle\lim_{x\to2}f\left(x\right)=
x
→
2
lim
f
(
x
)
=
c. i)
lim
x
→
3
−
f
(
x
)
=
\displaystyle\lim_{x\to3^-}f\left(x\right)=
x
→
3
−
lim
f
(
x
)
=
c. ii)
lim
x
→
3
+
f
(
x
)
=
\displaystyle\lim_{x\to3^+}f\left(x\right)=
x
→
3
+
lim
f
(
x
)
=
c. iii)
lim
x
→
3
f
(
x
)
=
\displaystyle\lim_{x\to3}f\left(x\right)=
x
→
3
lim
f
(
x
)
=
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: 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.
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∣