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Calculus Examples
limx→01+sin(4x)cot(x)limx→01+sin(4x)cot(x)
Step 1
Set up the limit as a left-sided limit.
limx→0-1+sin(4x)cot(x)limx→0−1+sin(4x)cot(x)
Step 2
Step 2.1
Evaluate the limit of 1+sin(4x)cot(x)1+sin(4x)cot(x) by plugging in 00 for xx.
1+sin(4⋅0)cot(0)1+sin(4⋅0)cot(0)
Step 2.2
Rewrite cot(0)cot(0) in terms of sines and cosines.
1+sin(4⋅0)cos(0)sin(0)1+sin(4⋅0)cos(0)sin(0)
Step 2.3
The exact value of sin(0)sin(0) is 00.
1+sin(4⋅0)cos(0)01+sin(4⋅0)cos(0)0
Step 2.4
Since 1+sin(4⋅0)cos(0)01+sin(4⋅0)cos(0)0 is undefined, the limit does not exist.
Does not existDoes not exist
Does not existDoes not exist
Step 3
Set up the limit as a right-sided limit.
limx→0+1+sin(4x)cot(x)limx→0+1+sin(4x)cot(x)
Step 4
Step 4.1
Evaluate the limit.
Step 4.1.1
Split the limit using the Sum of Limits Rule on the limit as xx approaches 0.
limx→0+1+limx→0+sin(4x)cot(x)
Step 4.1.2
Evaluate the limit of 1 which is constant as x approaches 0.
1+limx→0+sin(4x)cot(x)
1+limx→0+sin(4x)cot(x)
Step 4.2
Use the properties of logarithms to simplify the limit.
Step 4.2.1
Rewrite sin(4x)cot(x) as eln(sin(4x)cot(x)).
1+limx→0+eln(sin(4x)cot(x))
Step 4.2.2
Expand ln(sin(4x)cot(x)) by moving cot(x) outside the logarithm.
1+limx→0+ecot(x)ln(sin(4x))
1+limx→0+ecot(x)ln(sin(4x))
Step 4.3
Since the exponent cot(x)ln(sin(4x)) approaches -∞, the quantity ecot(x)ln(sin(4x)) approaches 0.
1+0
Step 4.4
Add 1 and 0.
1
1
Step 5
If either of the one-sided limits does not exist, the limit does not exist.
Does not exist