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Calculus Examples
limx→∞x-53√27x3+12limx→∞x−53√27x3+12
Step 1
Step 1.1
Factor 33 out of 27x327x3.
limx→∞x-53√3(9x3)+12limx→∞x−53√3(9x3)+12
Step 1.2
Factor 33 out of 1212.
limx→∞x-53√3(9x3)+3(4)limx→∞x−53√3(9x3)+3(4)
Step 1.3
Factor 33 out of 3(9x3)+3(4)3(9x3)+3(4).
limx→∞x-53√3(9x3+4)limx→∞x−53√3(9x3+4)
limx→∞x-53√3(9x3+4)limx→∞x−53√3(9x3+4)
Step 2
Divide the numerator and denominator by the highest power of xx in the denominator, which is x=3√x3x=3√x3.
limx→∞xx+-5x3√3(9x3+4)x3limx→∞xx+−5x3√3(9x3+4)x3
Step 3
Step 3.1
Simplify each term.
limx→∞1-5x3√3(9x3+4)x3limx→∞1−5x3√3(9x3+4)x3
Step 3.2
Split the limit using the Limits Quotient Rule on the limit as xx approaches ∞∞.
limx→∞1-5xlimx→∞3√3(9x3+4)x3limx→∞1−5xlimx→∞3√3(9x3+4)x3
Step 3.3
Split the limit using the Sum of Limits Rule on the limit as xx approaches ∞∞.
limx→∞1-limx→∞5xlimx→∞3√3(9x3+4)x3limx→∞1−limx→∞5xlimx→∞3√3(9x3+4)x3
Step 3.4
Evaluate the limit of 11 which is constant as xx approaches ∞∞.
1-limx→∞5xlimx→∞3√3(9x3+4)x31−limx→∞5xlimx→∞3√3(9x3+4)x3
Step 3.5
Move the term 55 outside of the limit because it is constant with respect to xx.
1-5limx→∞1xlimx→∞3√3(9x3+4)x31−5limx→∞1xlimx→∞3√3(9x3+4)x3
1-5limx→∞1xlimx→∞3√3(9x3+4)x31−5limx→∞1xlimx→∞3√3(9x3+4)x3
Step 4
Since its numerator approaches a real number while its denominator is unbounded, the fraction 1x approaches 0.
1-5⋅0limx→∞3√3(9x3+4)x3
Step 5
Step 5.1
Move the limit under the radical sign.
1-5⋅03√limx→∞3(9x3+4)x3
Step 5.2
Move the term 3 outside of the limit because it is constant with respect to x.
1-5⋅03√3limx→∞9x3+4x3
1-5⋅03√3limx→∞9x3+4x3
Step 6
Divide the numerator and denominator by the highest power of x in the denominator, which is x3.
1-5⋅03√3limx→∞9x3x3+4x3x3x3
Step 7
Step 7.1
Cancel the common factor of x3.
Step 7.1.1
Cancel the common factor.
1-5⋅03√3limx→∞9x3x3+4x3x3x3
Step 7.1.2
Divide 9 by 1.
1-5⋅03√3limx→∞9+4x3x3x3
1-5⋅03√3limx→∞9+4x3x3x3
Step 7.2
Cancel the common factor of x3.
Step 7.2.1
Cancel the common factor.
1-5⋅03√3limx→∞9+4x3x3x3
Step 7.2.2
Rewrite the expression.
1-5⋅03√3limx→∞9+4x31
1-5⋅03√3limx→∞9+4x31
Step 7.3
Split the limit using the Limits Quotient Rule on the limit as x approaches ∞.
1-5⋅03√3limx→∞9+4x3limx→∞1
Step 7.4
Split the limit using the Sum of Limits Rule on the limit as x approaches ∞.
1-5⋅03√3limx→∞9+limx→∞4x3limx→∞1
Step 7.5
Evaluate the limit of 9 which is constant as x approaches ∞.
1-5⋅03√39+limx→∞4x3limx→∞1
Step 7.6
Move the term 4 outside of the limit because it is constant with respect to x.
1-5⋅03√39+4limx→∞1x3limx→∞1
1-5⋅03√39+4limx→∞1x3limx→∞1
Step 8
Since its numerator approaches a real number while its denominator is unbounded, the fraction 1x3 approaches 0.
1-5⋅03√39+4⋅0limx→∞1
Step 9
Step 9.1
Evaluate the limit of 1 which is constant as x approaches ∞.
1-5⋅03√39+4⋅01
Step 9.2
Simplify the answer.
Step 9.2.1
Divide 9+4⋅0 by 1.
1-5⋅03√3(9+4⋅0)
Step 9.2.2
Simplify the numerator.
Step 9.2.2.1
Multiply -5 by 0.
1+03√3(9+4⋅0)
Step 9.2.2.2
Add 1 and 0.
13√3(9+4⋅0)
13√3(9+4⋅0)
Step 9.2.3
Simplify the denominator.
Step 9.2.3.1
Multiply 4 by 0.
13√3(9+0)
Step 9.2.3.2
Add 9 and 0.
13√3⋅9
Step 9.2.3.3
Multiply 3 by 9.
13√27
Step 9.2.3.4
Rewrite 27 as 33.
13√33
Step 9.2.3.5
Pull terms out from under the radical, assuming real numbers.
13
13
13
13
Step 10
The result can be shown in multiple forms.
Exact Form:
13
Decimal Form:
0.‾3