Calculus Examples

Evaluate the Limit limit as x approaches 0 of (sin(5x))/(3x)
limx0sin(5x)3xlimx0sin(5x)3x
Step 1
Move the term 1313 outside of the limit because it is constant with respect to xx.
13limx0sin(5x)x13limx0sin(5x)x
Step 2
Apply L'Hospital's rule.
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Step 2.1
Evaluate the limit of the numerator and the limit of the denominator.
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Step 2.1.1
Take the limit of the numerator and the limit of the denominator.
13limx0sin(5x)limx0x13limx0sin(5x)limx0x
Step 2.1.2
Evaluate the limit of the numerator.
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Step 2.1.2.1
Evaluate the limit.
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Step 2.1.2.1.1
Move the limit inside the trig function because sine is continuous.
13sin(limx05x)limx0x13sin(limx05x)limx0x
Step 2.1.2.1.2
Move the term 55 outside of the limit because it is constant with respect to xx.
13sin(5limx0x)limx0x13sin(5limx0x)limx0x
13sin(5limx0x)limx0x13sin(5limx0x)limx0x
Step 2.1.2.2
Evaluate the limit of xx by plugging in 00 for xx.
13sin(50)limx0x13sin(50)limx0x
Step 2.1.2.3
Simplify the answer.
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Step 2.1.2.3.1
Multiply 55 by 00.
13sin(0)limx0x13sin(0)limx0x
Step 2.1.2.3.2
The exact value of sin(0)sin(0) is 00.
130limx0x130limx0x
130limx0x130limx0x
130limx0x130limx0x
Step 2.1.3
Evaluate the limit of xx by plugging in 00 for xx.
13001300
Step 2.1.4
The expression contains a division by 00. The expression is undefined.
Undefined
13001300
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.
limx0sin(5x)x=limx0ddx[sin(5x)]ddx[x]limx0sin(5x)x=limx0ddx[sin(5x)]ddx[x]
Step 2.3
Find the derivative of the numerator and denominator.
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Step 2.3.1
Differentiate the numerator and denominator.
13limx0ddx[sin(5x)]ddx[x]13limx0ddx[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.
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Step 2.3.2.1
To apply the Chain Rule, set u as 5x.
13limx0ddu[sin(u)]ddx[5x]ddx[x]
Step 2.3.2.2
The derivative of sin(u) with respect to u is cos(u).
13limx0cos(u)ddx[5x]ddx[x]
Step 2.3.2.3
Replace all occurrences of u with 5x.
13limx0cos(5x)ddx[5x]ddx[x]
13limx0cos(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].
13limx0cos(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.
13limx0cos(5x)51ddx[x]
Step 2.3.5
Multiply 5 by 1.
13limx0cos(5x)5ddx[x]
Step 2.3.6
Move 5 to the left of cos(5x).
13limx05cos(5x)ddx[x]
Step 2.3.7
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
13limx05cos(5x)1
13limx05cos(5x)1
Step 2.4
Divide 5cos(5x) by 1.
13limx05cos(5x)
13limx05cos(5x)
Step 3
Evaluate the limit.
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Step 3.1
Move the term 5 outside of the limit because it is constant with respect to x.
135limx0cos(5x)
Step 3.2
Move the limit inside the trig function because cosine is continuous.
135cos(limx05x)
Step 3.3
Move the term 5 outside of the limit because it is constant with respect to x.
135cos(5limx0x)
135cos(5limx0x)
Step 4
Evaluate the limit of x by plugging in 0 for x.
135cos(50)
Step 5
Simplify the answer.
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Step 5.1
Combine 13 and 5.
53cos(50)
Step 5.2
Multiply 5 by 0.
53cos(0)
Step 5.3
The exact value of cos(0) is 1.
531
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
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