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Math Problems
Algebra 2
Power rule
Let
f
(
x
)
=
−
2
x
2
+
5
x
4
x
3
−
2
x
+
1
f(x)=\frac{-2 x^{2}+5 x}{4 x^{3}-2 x+1}
f
(
x
)
=
4
x
3
−
2
x
+
1
−
2
x
2
+
5
x
.
\newline
Find
lim
x
→
∞
f
(
x
)
\lim _{x \rightarrow \infty} f(x)
lim
x
→
∞
f
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
0
0
0
\newline
(B)
5
5
5
\newline
(C)
−
1
2
-\frac{1}{2}
−
2
1
\newline
(D) The limit is unbounded
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Let
f
(
x
)
=
−
6
x
2
+
x
−
1
x
2
+
3
f(x)=\frac{-6 x^{2}+x-1}{x^{2}+3}
f
(
x
)
=
x
2
+
3
−
6
x
2
+
x
−
1
.
\newline
Find
lim
x
→
∞
f
(
x
)
\lim _{x \rightarrow \infty} f(x)
lim
x
→
∞
f
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
3
-\frac{1}{3}
−
3
1
\newline
(B)
−
6
-6
−
6
\newline
(C)
0
0
0
\newline
(D) The limit is unbounded
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h
(
x
)
=
3
x
−
11
x
2
−
1
h(x)=\frac{3 x-11}{x^{2}-1}
h
(
x
)
=
x
2
−
1
3
x
−
11
\newline
We want to find
lim
x
→
1
h
(
x
)
\lim _{x \rightarrow 1} h(x)
lim
x
→
1
h
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
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h
(
x
)
=
54
+
x
−
7
6
x
+
30
h(x)=\frac{\sqrt{54+x}-7}{6 x+30}
h
(
x
)
=
6
x
+
30
54
+
x
−
7
\newline
We want to find
lim
x
→
−
5
h
(
x
)
\lim _{x \rightarrow-5} h(x)
lim
x
→
−
5
h
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
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h
(
x
)
=
cot
2
(
x
)
1
+
2
sin
(
x
)
h(x)=\frac{\cot ^{2}(x)}{1+\sqrt{2} \sin (x)}
h
(
x
)
=
1
+
2
sin
(
x
)
cot
2
(
x
)
\newline
We want to find
lim
x
→
−
π
4
h
(
x
)
\lim _{x \rightarrow-\frac{\pi}{4}} h(x)
lim
x
→
−
4
π
h
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
f
(
x
)
=
x
+
2
x
2
−
3
x
+
2
f(x)=\frac{x+2}{x^{2}-3 x+2}
f
(
x
)
=
x
2
−
3
x
+
2
x
+
2
\newline
We want to find
lim
x
→
3
f
(
x
)
\lim _{x \rightarrow 3} f(x)
lim
x
→
3
f
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
h
(
x
)
=
x
2
−
6
x
+
10
x
−
1
h(x)=\frac{x^{2}-6 x+10}{x-1}
h
(
x
)
=
x
−
1
x
2
−
6
x
+
10
\newline
We want to find
lim
x
→
2
h
(
x
)
\lim _{x \rightarrow 2} h(x)
lim
x
→
2
h
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
f
(
x
)
=
2
−
cos
2
(
x
)
sin
(
2
x
)
f(x)=\frac{2-\cos ^{2}(x)}{\sin (2 x)}
f
(
x
)
=
sin
(
2
x
)
2
−
cos
2
(
x
)
\newline
We want to find
lim
x
→
π
2
f
(
x
)
\lim _{x \rightarrow \frac{\pi}{2}} f(x)
lim
x
→
2
π
f
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
f
(
x
)
=
3
x
−
15
−
3
x
−
8
f(x)=\frac{\sqrt{3 x-15}-3}{x-8}
f
(
x
)
=
x
−
8
3
x
−
15
−
3
\newline
We want to find
lim
x
→
8
f
(
x
)
\lim _{x \rightarrow 8} f(x)
lim
x
→
8
f
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
h
(
x
)
=
7
−
4
x
3
+
9
x
−
2
h(x)=\frac{7-4 x}{3+\sqrt{9 x-2}}
h
(
x
)
=
3
+
9
x
−
2
7
−
4
x
\newline
We want to find
lim
x
→
3
h
(
x
)
\lim _{x \rightarrow 3} h(x)
lim
x
→
3
h
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
f
(
x
)
=
4
x
+
4
x
3
+
6
x
2
+
5
x
f(x)=\frac{4 x+4}{x^{3}+6 x^{2}+5 x}
f
(
x
)
=
x
3
+
6
x
2
+
5
x
4
x
+
4
\newline
We want to find
lim
x
→
−
1
f
(
x
)
\lim _{x \rightarrow-1} f(x)
lim
x
→
−
1
f
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
g
(
x
)
=
cos
(
x
)
−
sin
(
x
)
cos
(
2
x
)
g(x)=\frac{\cos (x)-\sin (x)}{\cos (2 x)}
g
(
x
)
=
cos
(
2
x
)
cos
(
x
)
−
sin
(
x
)
\newline
We want to find
lim
x
→
π
4
g
(
x
)
\lim _{x \rightarrow \frac{\pi}{4}} g(x)
lim
x
→
4
π
g
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
g
(
x
)
=
tan
(
x
)
1
−
cos
(
x
)
g(x)=\frac{\tan (x)}{1-\cos (x)}
g
(
x
)
=
1
−
cos
(
x
)
tan
(
x
)
\newline
We want to find
lim
x
→
π
g
(
x
)
\lim _{x \rightarrow \pi} g(x)
lim
x
→
π
g
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
Get tutor help
Simplify. Assume all variables are positive.
\newline
(
27
x
)
2
3
(27x)^{\frac{2}{3}}
(
27
x
)
3
2
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
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