Calculus Examples

Evaluate the Limit limit as x approaches 0 of (sin(5x))/(5x)
limx0sin(5x)5x
Step 1
Evaluate the limit of the numerator and the limit of the denominator.
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Step 1.1
Take the limit of the numerator and the limit of the denominator.
limx0sin(5x)limx05x
Step 1.2
Evaluate the limit of the numerator.
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Step 1.2.1
Evaluate the limit.
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Step 1.2.1.1
Move the limit inside the trig function because sine is continuous.
sin(limx05x)limx05x
Step 1.2.1.2
Move the term 5 outside of the limit because it is constant with respect to x.
sin(5limx0x)limx05x
sin(5limx0x)limx05x
Step 1.2.2
Evaluate the limit of x by plugging in 0 for x.
sin(50)limx05x
Step 1.2.3
Simplify the answer.
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Step 1.2.3.1
Multiply 5 by 0.
sin(0)limx05x
Step 1.2.3.2
The exact value of sin(0) is 0.
0limx05x
0limx05x
0limx05x
Step 1.3
Evaluate the limit of the denominator.
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Step 1.3.1
Move the term 5 outside of the limit because it is constant with respect to x.
05limx0x
Step 1.3.2
Evaluate the limit of x by plugging in 0 for x.
050
Step 1.3.3
Multiply 5 by 0.
00
Step 1.3.4
The expression contains a division by 0. The expression is undefined.
Undefined
00
Step 1.4
The expression contains a division by 0. The expression is undefined.
Undefined
00
Step 2
Since 00 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.
limx0sin(5x)5x=limx0ddx[sin(5x)]ddx[5x]
Step 3
Find the derivative of the numerator and denominator.
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Step 3.1
Differentiate the numerator and denominator.
limx0ddx[sin(5x)]ddx[5x]
Step 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)=5x.
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Step 3.2.1
To apply the Chain Rule, set u as 5x.
limx0ddu[sin(u)]ddx[5x]ddx[5x]
Step 3.2.2
The derivative of sin(u) with respect to u is cos(u).
limx0cos(u)ddx[5x]ddx[5x]
Step 3.2.3
Replace all occurrences of u with 5x.
limx0cos(5x)ddx[5x]ddx[5x]
limx0cos(5x)ddx[5x]ddx[5x]
Step 3.3
Since 5 is constant with respect to x, the derivative of 5x with respect to x is 5ddx[x].
limx0cos(5x)(5ddx[x])ddx[5x]
Step 3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx0cos(5x)(51)ddx[5x]
Step 3.5
Multiply 5 by 1.
limx0cos(5x)5ddx[5x]
Step 3.6
Move 5 to the left of cos(5x).
limx05cos(5x)ddx[5x]
Step 3.7
Since 5 is constant with respect to x, the derivative of 5x with respect to x is 5ddx[x].
limx05cos(5x)5ddx[x]
Step 3.8
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx05cos(5x)51
Step 3.9
Multiply 5 by 1.
limx05cos(5x)5
limx05cos(5x)5
Step 4
Evaluate the limit.
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Step 4.1
Cancel the common factor of 5.
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Step 4.1.1
Cancel the common factor.
limx05cos(5x)5
Step 4.1.2
Divide cos(5x) by 1.
limx0cos(5x)
limx0cos(5x)
Step 4.2
Move the limit inside the trig function because cosine is continuous.
cos(limx05x)
Step 4.3
Move the term 5 outside of the limit because it is constant with respect to x.
cos(5limx0x)
cos(5limx0x)
Step 5
Evaluate the limit of x by plugging in 0 for x.
cos(50)
Step 6
Simplify the answer.
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Step 6.1
Multiply 5 by 0.
cos(0)
Step 6.2
The exact value of cos(0) is 1.
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limx0(sin(5x)5x)
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