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
Limits by Multiplying by the Conjugate
Related Topics
Wize University Calculus 1 Textbook > Limits
Computing Limits by Multiplying by the Conjugate
3 Activities
lim
x
→
0
x
2
+
x
−
2
−
x
\displaystyle\lim_{x\rightarrow0}\dfrac{x}{\sqrt{2+x}-\sqrt{2-x}}
x
→
0
lim
2
+
x
−
2
−
x
x
2
2
2\sqrt{2}
2
2
2
2
2
2
\sqrt{2}
2
1
1
1
I don't know
Check Submission
More Computing Limits by Multiplying by the Conjugate Questions:
Limits: Multiplying by the Conjugate
lim
x
→
∞
x
2
+
2
x
−
x
2
+
1
\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
lim
x
→
∞
x
2
+
2
x
−
x
2
+
1
Limit Techniques: Multiply by the Conjugate
lim
x
→
π
2
1
−
sin
x
cos
x
\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin{x}}{\cos{x}}
x
→
2
π
lim
cos
x
1
−
sin
x
∞ - ∞ Limits with Square Root
lim
x
→
∞
x
2
+
2
x
−
x
2
−
3
x
=
\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
x
→
∞
lim
x
2
+
2
x
−
x
2
−
3
x
=
∞ - ∞ Limits with Square Root
lim
x
→
∞
x
2
+
2
x
−
x
2
−
3
x
=
\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
x
→
∞
lim
x
2
+
2
x
−
x
2
−
3
x
=
Evaluating limits: Multiplying by the Conjugate
Practice: Evaluating limits
Determine the value of
lim
x
→
0
x
+
1
−
1
−
x
x
\lim_{x\ \rightarrow0}\ \frac{\sqrt{x+1}-\sqrt{1-x}}{x}\
lim
x
→
0
x
x
+
1
−
1
−
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
Practice: Limit at Infinity with Roots
Q.
\textbf{Q.}
Q.
Evaluate the limit:
lim
x
→
∞
x
2
−
2
−
x
2
+
x
\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2-2}-\sqrt{x^2+x}
x
→
∞
lim
x
2
−
2
−
x
2
+
x
Limit at Infinity with Roots
Evaluate
lim
x
→
∞
9
x
2
+
2
x
−
3
x
\displaystyle \lim_{x\rightarrow\infty}\sqrt{9x^2+2x}-3x
x
→
∞
lim
9
x
2
+
2
x
−
3
x
.
Limit with Unknown Coefficients
Find constants
a
a
a
and
b
b
b
such that
lim
x
→
0
a
x
+
b
−
2
x
=
1
\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1
x
→
0
lim
x
a
x
+
b
−
2
=
1
.
Square Roots: Limits by Multiplying by the Conjugate
Q.
\textbf{Q.}
Q.
Evaluate
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
Limits by multiplying by the conjugate
lim
t
→
1
3
+
t
−
2
t
3
−
1
\lim_{t\rightarrow1}\dfrac{\sqrt{3+t}-2}{t^3-1}
lim
t
→
1
t
3
−
1
3
+
t
−
2
Evaluate
lim
x
→
9
x
+
7
−
4
2
x
−
18
\displaystyle\lim_{x\to9} \frac{\sqrt{x+7}-4}{2x-18}
x
→
9
lim
2
x
−
18
x
+
7
−
4
l'Hopital's Rule
Evaluate
lim
x
→
∞
x
2
+
1
−
x
\lim\limits_{x\to\infty} \sqrt{x^2+1}-x
x
→
∞
lim
x
2
+
1
−
x
Practice: Rationalizing Indeterminate Forms
Rationalizing: Evaluate
lim
x
→
3
x
2
−
9
x
−
3
\lim_{x\to3}\limits\dfrac{x^2 - 9}{\sqrt x - \sqrt 3}
x
→
3
lim
x
−
3
x
2
−
9
Limits: Multiplying Conjugates
Find constants
a
and
b
such that
lim
x
→
0
a
x
+
b
−
2
x
=
1
\lim\limits_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1
x
→
0
lim
x
a
x
+
b
−
2
=
1
Limits: Multiplying Conjugates
Evaluate the limit:
lim
x
→
2
x
+
7
−
3
x
−
2
\lim\limits_{x \to 2} \frac{\sqrt{x+7} - 3}{x-2}
x
→
2
lim
x
−
2
x
+
7
−
3
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 by Multiplying by the Conjugate
The value of the limits
lim
x
→
1
x
−
1
x
2
−
4
x
+
3
\lim\limits_{x\rightarrow1}\frac{\sqrt{x}-1}{x^2-4x+3}
x
→
1
lim
x
2
−
4
x
+
3
x
−
1
Determine the value of
lim
x
→
2
4
x
−
1
−
7
x
−
2
\displaystyle \lim_{x \to 2} \frac{\sqrt{4x-1}-\sqrt{7}}{x-2}
x
→
2
lim
x
−
2
4
x
−
1
−
7
.
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
Limit with Unknown Coefficients
Find constants
a
a
a
and
b
b
b
such that
lim
x
→
0
a
x
+
b
−
2
x
=
1
\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1
x
→
0
lim
x
a
x
+
b
−
2
=
1
.
Square Roots: Limits by Multiplying by the Conjugate
Q.
\textbf{Q.}
Q.
Evaluate
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
Practice: Limit at Infinity with Roots
Q.
\textbf{Q.}
Q.
Evaluate the limit:
lim
x
→
∞
x
2
−
2
−
x
2
+
x
\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2-2}-\sqrt{x^2+x}
x
→
∞
lim
x
2
−
2
−
x
2
+
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
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
Limit at Infinity with Roots
Evaluate
lim
x
→
∞
9
x
2
+
2
x
−
3
x
\displaystyle \lim_{x\rightarrow\infty}\sqrt{9x^2+2x}-3x
x
→
∞
lim
9
x
2
+
2
x
−
3
x
.
Practice: Multiply by conjugate
Evaluate
lim
x
→
−
∞
(
x
2
+
3
x
+
x
)
\lim\limits_{x\rightarrow-\infin}(\sqrt{x^2+3x}+x)
x
→
−
∞
lim
(
x
2
+
3
x
+
x
)
Evaluate the limit
lim
x
→
2
3
x
−
1
−
5
x
−
2
\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{3x-1}-\sqrt{5}}{x-2}
x
→
2
lim
x
−
2
3
x
−
1
−
5
Evaluate
lim
t
→
1
3
+
t
−
2
t
3
−
1
\displaystyle \lim_{t\rightarrow 1}\frac{\sqrt{3+t}-2}{t^3-1}
t
→
1
lim
t
3
−
1
3
+
t
−
2
.
Determine the value of
lim
x
→
∞
4
x
4
+
4
x
2
+
1
−
2
x
2
\displaystyle \lim_{x\rightarrow \infty} \sqrt{4x^4+4x^2+1}-2x^2
x
→
∞
lim
4
x
4
+
4
x
2
+
1
−
2
x
2
.
🌶️
TOUGH!
Evaluate the limit,
lim
x
→
−
∞
x
2
−
6
x
+
x
\displaystyle\lim_{x\rightarrow-\infty}\sqrt{x^2-6x}+x
x
→
−
∞
lim
x
2
−
6
x
+
x
.
Find the following limit:
lim
x
→
+
∞
x
2
−
8
x
−
x
\displaystyle \lim_{x\rightarrow+\infty}\sqrt{x^2-8x}-x
x
→
+
∞
lim
x
2
−
8
x
−
x
.
Evaluate
lim
x
→
5
x
−
5
x
2
−
4
x
−
5
\displaystyle \lim_{x\rightarrow 5}\frac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}
x
→
5
lim
x
2
−
4
x
−
5
x
−
5
.
Limits: Combining techniques
Solve the limit
lim
x
→
3
9
x
−
6
x
−
1
x
2
−
5
−
2
\displaystyle \lim\limits_{x\to 3} \frac{\frac{9}{x}-\frac{6}{x-1}}{\sqrt{x^2-5}-2}
x
→
3
lim
x
2
−
5
−
2
x
9
−
x
−
1
6
.
lim
x
→
5
x
−
5
x
2
−
4
x
−
5
\displaystyle\lim_{x\rightarrow5}\dfrac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}
x
→
5
lim
x
2
−
4
x
−
5
x
−
5
lim
x
→
∞
x
2
+
2
x
−
x
2
+
1
\displaystyle\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
x
→
∞
lim
x
2
+
2
x
−
x
2
+
1
Evaluating limits: Multiplying by the Conjugate
Practice: Evaluating limits
Determine the value of
lim
x
→
0
x
+
1
−
1
−
x
x
\lim_{x\ \rightarrow0}\ \frac{\sqrt{x+1}-\sqrt{1-x}}{x}\
lim
x
→
0
x
x
+
1
−
1
−
x
∞ - ∞ Limits with Square Root
Practice: ∞ - ∞ Limits with Square Root
lim
x
→
∞
x
2
+
2
x
−
x
2
−
3
x
=
\displaystyle\lim_{x\to\infty}\ \sqrt{x^2+2x}-\sqrt{x^2-3x}=
x
→
∞
lim
x
2
+
2
x
−
x
2
−
3
x
=
Practice: $\frac{0}{0}$ Limit with Square Root
Practice: 0/0 Limit with Square Root
lim
x
→
0
x
x
+
4
−
4
−
x
=
\displaystyle\lim_{x\to0}\ \frac{x}{\sqrt{x+4}-\sqrt{4-x}}=
x
→
0
lim
x
+
4
−
4
−
x
x
=
Limits and Continuity
Evaluate the following limits:
a)
lim
x
→
4
x
−
4
x
−
4
\displaystyle \lim_{x \rightarrow 4} \frac{\sqrt{x} - \sqrt{4}}{x - 4}
x
→
4
lim
x
−
4
x
−
4
b)
lim
x
→
∞
x
2
−
5
x
+
13
4
x
4
+
3
x
−
21
\displaystyle \lim_{x \rightarrow \infty} \frac{x^2 - 5x + 13}{\sqrt{4x^4 +3x - 21}}
x
→
∞
lim
4
x
4
+
3
x
−
21
x
2
−
5
x
+
13
Limit Techniques: Multiply by the Conjugate
lim
x
→
π
2
1
−
sin
x
cos
x
\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin{x}}{\cos{x}}
x
→
2
π
lim
cos
x
1
−
sin
x
Limits by multiplying by the conjugate
lim
x
→
−
∞
(
9
x
2
−
15
x
+
3
+
3
x
)
\displaystyle\lim_{x\rightarrow -\infty}\big(\sqrt{9x^2-15x+3}+3x\big)
x
→
−
∞
lim
(
9
x
2
−
15
x
+
3
+
3
x
)
Limits by multiplying by the conjugate
lim
x
→
5
x
−
5
x
2
−
4
x
−
5
\displaystyle\lim_{x\rightarrow5}\dfrac{\sqrt{x}-\sqrt{5}}{x^2-4x-5}
x
→
5
lim
x
2
−
4
x
−
5
x
−
5
Limits: Multiplying by the Conjugate
lim
x
→
∞
x
2
+
2
x
−
x
2
+
1
\lim_{x\rightarrow\infty}\sqrt{x^2+2x}-\sqrt{x^2+1}
lim
x
→
∞
x
2
+
2
x
−
x
2
+
1
Practice: L'hopital's rule
Q.
\textbf{Q.}
Q.
Evaluate the following limits.
1.
lim
x
→
0
+
cos
x
cot
x
\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
x
→
0
+
lim
cos
x
c
o
t
x
2.
lim
x
→
∞
[
x
2
+
1
−
x
]
\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt{x^2+1}-x\right]
x
→
∞
lim
[
x
2
+
1
−
x
]