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Calculus Examples
limx→0sin(3x)xlimx→0sin(3x)x
Step 1
Step 1.1
Evaluate the limit of the numerator and the limit of the denominator.
Step 1.1.1
Take the limit of the numerator and the limit of the denominator.
limx→0sin(3x)limx→0xlimx→0sin(3x)limx→0x
Step 1.1.2
Evaluate the limit of the numerator.
Step 1.1.2.1
Evaluate the limit.
Step 1.1.2.1.1
Move the limit inside the trig function because sine is continuous.
sin(limx→03x)limx→0xsin(limx→03x)limx→0x
Step 1.1.2.1.2
Move the term 33 outside of the limit because it is constant with respect to xx.
sin(3limx→0x)limx→0xsin(3limx→0x)limx→0x
sin(3limx→0x)limx→0xsin(3limx→0x)limx→0x
Step 1.1.2.2
Evaluate the limit of xx by plugging in 00 for xx.
sin(3⋅0)limx→0xsin(3⋅0)limx→0x
Step 1.1.2.3
Simplify the answer.
Step 1.1.2.3.1
Multiply 33 by 00.
sin(0)limx→0xsin(0)limx→0x
Step 1.1.2.3.2
The exact value of sin(0)sin(0) is 00.
0limx→0x0limx→0x
0limx→0x0limx→0x
0limx→0x0limx→0x
Step 1.1.3
Evaluate the limit of xx by plugging in 00 for xx.
0000
Step 1.1.4
The expression contains a division by 00. The expression is undefined.
Undefined
0000
Step 1.2
Since 0000 is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
limx→0sin(3x)x=limx→0ddx[sin(3x)]ddx[x]limx→0sin(3x)x=limx→0ddx[sin(3x)]ddx[x]
Step 1.3
Find the derivative of the numerator and denominator.
Step 1.3.1
Differentiate the numerator and denominator.
limx→0ddx[sin(3x)]ddx[x]
Step 1.3.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=sin(x) and g(x)=3x.
Step 1.3.2.1
To apply the Chain Rule, set u as 3x.
limx→0ddu[sin(u)]ddx[3x]ddx[x]
Step 1.3.2.2
The derivative of sin(u) with respect to u is cos(u).
limx→0cos(u)ddx[3x]ddx[x]
Step 1.3.2.3
Replace all occurrences of u with 3x.
limx→0cos(3x)ddx[3x]ddx[x]
limx→0cos(3x)ddx[3x]ddx[x]
Step 1.3.3
Since 3 is constant with respect to x, the derivative of 3x with respect to x is 3ddx[x].
limx→0cos(3x)(3ddx[x])ddx[x]
Step 1.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→0cos(3x)(3⋅1)ddx[x]
Step 1.3.5
Multiply 3 by 1.
limx→0cos(3x)⋅3ddx[x]
Step 1.3.6
Move 3 to the left of cos(3x).
limx→03⋅cos(3x)ddx[x]
Step 1.3.7
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→03cos(3x)1
limx→03cos(3x)1
Step 1.4
Divide 3cos(3x) by 1.
limx→03cos(3x)
limx→03cos(3x)
Step 2
Step 2.1
Move the term 3 outside of the limit because it is constant with respect to x.
3limx→0cos(3x)
Step 2.2
Move the limit inside the trig function because cosine is continuous.
3cos(limx→03x)
Step 2.3
Move the term 3 outside of the limit because it is constant with respect to x.
3cos(3limx→0x)
3cos(3limx→0x)
Step 3
Evaluate the limit of x by plugging in 0 for x.
3cos(3⋅0)
Step 4
Step 4.1
Multiply 3 by 0.
3cos(0)
Step 4.2
The exact value of cos(0) is 1.
3⋅1
Step 4.3
Multiply 3 by 1.
3
3