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Calculus Examples
limx→0sin(5x)3xlimx→0sin(5x)3x
Step 1
Move the term 1313 outside of the limit because it is constant with respect to xx.
13limx→0sin(5x)x13limx→0sin(5x)x
Step 2
Step 2.1
Evaluate the limit of the numerator and the limit of the denominator.
Step 2.1.1
Take the limit of the numerator and the limit of the denominator.
13⋅limx→0sin(5x)limx→0x13⋅limx→0sin(5x)limx→0x
Step 2.1.2
Evaluate the limit of the numerator.
Step 2.1.2.1
Evaluate the limit.
Step 2.1.2.1.1
Move the limit inside the trig function because sine is continuous.
13⋅sin(limx→05x)limx→0x13⋅sin(limx→05x)limx→0x
Step 2.1.2.1.2
Move the term 55 outside of the limit because it is constant with respect to xx.
13⋅sin(5limx→0x)limx→0x13⋅sin(5limx→0x)limx→0x
13⋅sin(5limx→0x)limx→0x13⋅sin(5limx→0x)limx→0x
Step 2.1.2.2
Evaluate the limit of xx by plugging in 00 for xx.
13⋅sin(5⋅0)limx→0x13⋅sin(5⋅0)limx→0x
Step 2.1.2.3
Simplify the answer.
Step 2.1.2.3.1
Multiply 55 by 00.
13⋅sin(0)limx→0x13⋅sin(0)limx→0x
Step 2.1.2.3.2
The exact value of sin(0)sin(0) is 00.
13⋅0limx→0x13⋅0limx→0x
13⋅0limx→0x13⋅0limx→0x
13⋅0limx→0x13⋅0limx→0x
Step 2.1.3
Evaluate the limit of xx by plugging in 00 for xx.
13⋅0013⋅00
Step 2.1.4
The expression contains a division by 00. The expression is undefined.
Undefined
13⋅0013⋅00
Step 2.2
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(5x)x=limx→0ddx[sin(5x)]ddx[x]limx→0sin(5x)x=limx→0ddx[sin(5x)]ddx[x]
Step 2.3
Find the derivative of the numerator and denominator.
Step 2.3.1
Differentiate the numerator and denominator.
13limx→0ddx[sin(5x)]ddx[x]13limx→0ddx[sin(5x)]ddx[x]
Step 2.3.2
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=sin(x) and g(x)=5x.
Step 2.3.2.1
To apply the Chain Rule, set u as 5x.
13limx→0ddu[sin(u)]ddx[5x]ddx[x]
Step 2.3.2.2
The derivative of sin(u) with respect to u is cos(u).
13limx→0cos(u)ddx[5x]ddx[x]
Step 2.3.2.3
Replace all occurrences of u with 5x.
13limx→0cos(5x)ddx[5x]ddx[x]
13limx→0cos(5x)ddx[5x]ddx[x]
Step 2.3.3
Since 5 is constant with respect to x, the derivative of 5x with respect to x is 5ddx[x].
13limx→0cos(5x)⋅5ddx[x]ddx[x]
Step 2.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
13limx→0cos(5x)⋅5⋅1ddx[x]
Step 2.3.5
Multiply 5 by 1.
13limx→0cos(5x)⋅5ddx[x]
Step 2.3.6
Move 5 to the left of cos(5x).
13limx→05⋅cos(5x)ddx[x]
Step 2.3.7
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
13limx→05cos(5x)1
13limx→05cos(5x)1
Step 2.4
Divide 5cos(5x) by 1.
13limx→05cos(5x)
13limx→05cos(5x)
Step 3
Step 3.1
Move the term 5 outside of the limit because it is constant with respect to x.
13⋅5limx→0cos(5x)
Step 3.2
Move the limit inside the trig function because cosine is continuous.
13⋅5cos(limx→05x)
Step 3.3
Move the term 5 outside of the limit because it is constant with respect to x.
13⋅5cos(5limx→0x)
13⋅5cos(5limx→0x)
Step 4
Evaluate the limit of x by plugging in 0 for x.
13⋅5cos(5⋅0)
Step 5
Step 5.1
Combine 13 and 5.
53cos(5⋅0)
Step 5.2
Multiply 5 by 0.
53cos(0)
Step 5.3
The exact value of cos(0) is 1.
53⋅1
Step 5.4
Multiply 53 by 1.
53
53
Step 6
The result can be shown in multiple forms.
Exact Form:
53
Decimal Form:
1.‾6