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Calculus Examples
limx→π3cot(-2√3xπ+csc(x)-π4)limx→π3cot(−2√3xπ+csc(x)−π4)
Step 1
Move the limit inside the trig function because cotangent is continuous.
cot(limx→π3-2√3xπ+csc(x)-π4)cot(limx→π3−2√3xπ+csc(x)−π4)
Step 2
Split the limit using the Sum of Limits Rule on the limit as xx approaches π3π3.
cot(-limx→π32√3xπ+limx→π3csc(x)-limx→π3π4)cot(−limx→π32√3xπ+limx→π3csc(x)−limx→π3π4)
Step 3
Move the term 2√3π2√3π outside of the limit because it is constant with respect to xx.
cot(-(2√3πlimx→π3x)+limx→π3csc(x)-limx→π3π4)cot(−(2√3πlimx→π3x)+limx→π3csc(x)−limx→π3π4)
Step 4
Move the limit inside the trig function because cosecant is continuous.
cot(-(2√3πlimx→π3x)+csc(limx→π3x)-limx→π3π4)cot(−(2√3πlimx→π3x)+csc(limx→π3x)−limx→π3π4)
Step 5
Evaluate the limit of π4π4 which is constant as xx approaches π3π3.
cot(-(2√3πlimx→π3x)+csc(limx→π3x)-π4)cot(−(2√3πlimx→π3x)+csc(limx→π3x)−π4)
Step 6
Step 6.1
Evaluate the limit of xx by plugging in π3π3 for xx.
cot(-(2√3π⋅π3)+csc(limx→π3x)-π4)cot(−(2√3π⋅π3)+csc(limx→π3x)−π4)
Step 6.2
Evaluate the limit of xx by plugging in π3π3 for xx.
cot(-(2√3π⋅π3)+csc(π3)-π4)cot(−(2√3π⋅π3)+csc(π3)−π4)
cot(-(2√3π⋅π3)+csc(π3)-π4)cot(−(2√3π⋅π3)+csc(π3)−π4)
Step 7
Step 7.1
Simplify each term.
Step 7.1.1
Cancel the common factor of ππ.
Step 7.1.1.1
Cancel the common factor.
cot(-(2√3π⋅π3)+csc(π3)-π4)
Step 7.1.1.2
Rewrite the expression.
cot(-(2√313)+csc(π3)-π4)
cot(-(2√313)+csc(π3)-π4)
Step 7.1.2
Combine 13 and 2.
cot(-(23√3)+csc(π3)-π4)
Step 7.1.3
Combine 23 and √3.
cot(-2√33+csc(π3)-π4)
Step 7.1.4
The exact value of csc(π3) is 2√3.
cot(-2√33+2√3-π4)
Step 7.1.5
Multiply 2√3 by √3√3.
cot(-2√33+2√3⋅√3√3-π4)
Step 7.1.6
Combine and simplify the denominator.
Step 7.1.6.1
Multiply 2√3 by √3√3.
cot(-2√33+2√3√3√3-π4)
Step 7.1.6.2
Raise √3 to the power of 1.
cot(-2√33+2√3√31√3-π4)
Step 7.1.6.3
Raise √3 to the power of 1.
cot(-2√33+2√3√31√31-π4)
Step 7.1.6.4
Use the power rule aman=am+n to combine exponents.
cot(-2√33+2√3√31+1-π4)
Step 7.1.6.5
Add 1 and 1.
cot(-2√33+2√3√32-π4)
Step 7.1.6.6
Rewrite √32 as 3.
Step 7.1.6.6.1
Use n√ax=axn to rewrite √3 as 312.
cot(-2√33+2√3(312)2-π4)
Step 7.1.6.6.2
Apply the power rule and multiply exponents, (am)n=amn.
cot(-2√33+2√3312⋅2-π4)
Step 7.1.6.6.3
Combine 12 and 2.
cot(-2√33+2√3322-π4)
Step 7.1.6.6.4
Cancel the common factor of 2.
Step 7.1.6.6.4.1
Cancel the common factor.
cot(-2√33+2√3322-π4)
Step 7.1.6.6.4.2
Rewrite the expression.
cot(-2√33+2√331-π4)
cot(-2√33+2√331-π4)
Step 7.1.6.6.5
Evaluate the exponent.
cot(-2√33+2√33-π4)
cot(-2√33+2√33-π4)
cot(-2√33+2√33-π4)
cot(-2√33+2√33-π4)
Step 7.2
Combine the numerators over the common denominator.
cot(-2√3+2√33+-π4)
Step 7.3
Add -2√3 and 2√3.
cot(03+-π4)
Step 7.4
Simplify each term.
Step 7.4.1
Divide 0 by 3.
cot(0+-π4)
Step 7.4.2
Move the negative in front of the fraction.
cot(0-π4)
cot(0-π4)
Step 7.5
Subtract π4 from 0.
cot(-π4)
Step 7.6
Add full rotations of 2π until the angle is greater than or equal to 0 and less than 2π.
cot(7π4)
Step 7.7
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the fourth quadrant.
-cot(π4)
Step 7.8
The exact value of cot(π4) is 1.
-1⋅1
Step 7.9
Multiply -1 by 1.
-1
-1