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Calculus Examples
limx→0arctan(2x)3x
Step 1
Move the term 13 outside of the limit because it is constant with respect to x.
13limx→0arctan(2x)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→0arctan(2x)limx→0x
Step 2.1.2
Evaluate the limit of the numerator.
Step 2.1.2.1
Evaluate the limit of x by plugging in 0 for x.
13⋅0limx→0x
Step 2.1.2.2
Substitute t for 2x and let t approach 0 since limx→02x=0.
13⋅limt→0arctan(t)limx→0x
Step 2.1.2.3
Evaluate the limits by plugging in 0 for all occurrences of t.
Step 2.1.2.3.1
Evaluate the limit of x by plugging in 0 for x.
13⋅arctan(0)limx→0x
Step 2.1.2.3.2
The exact value of arctan(0) is 0.
13⋅0limx→0x
13⋅0limx→0x
13⋅0limx→0x
Step 2.1.3
Evaluate the limit of x by plugging in 0 for x.
13⋅00
Step 2.1.4
The expression contains a division by 0. The expression is undefined.
Undefined
13⋅00
Step 2.2
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.
limx→0arctan(2x)x=limx→0ddx[arctan(2x)]ddx[x]
Step 2.3
Find the derivative of the numerator and denominator.
Step 2.3.1
Differentiate the numerator and denominator.
13limx→0ddx[arctan(2x)]ddx[x]
Step 2.3.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=arctan(x) and g(x)=2x.
Step 2.3.2.1
To apply the Chain Rule, set u as 2x.
13limx→0ddu[arctan(u)]ddx[2x]ddx[x]
Step 2.3.2.2
The derivative of arctan(u) with respect to u is 11+u2.
13limx→011+u2ddx[2x]ddx[x]
Step 2.3.2.3
Replace all occurrences of u with 2x.
13limx→011+(2x)2ddx[2x]ddx[x]
13limx→011+(2x)2ddx[2x]ddx[x]
Step 2.3.3
Factor 2 out of 2x.
13limx→011+(2(x))2ddx[2x]ddx[x]
Step 2.3.4
Apply the product rule to 2(x).
13limx→011+22x2ddx[2x]ddx[x]
Step 2.3.5
Raise 2 to the power of 2.
13limx→011+4x2ddx[2x]ddx[x]
Step 2.3.6
Since 2 is constant with respect to x, the derivative of 2x with respect to x is 2ddx[x].
13limx→011+4x2⋅2ddx[x]ddx[x]
Step 2.3.7
Combine 2 and 11+4x2.
13limx→021+4x2ddx[x]ddx[x]
Step 2.3.8
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
13limx→021+4x2⋅1ddx[x]
Step 2.3.9
Multiply 21+4x2 by 1.
13limx→021+4x2ddx[x]
Step 2.3.10
Reorder terms.
13limx→024x2+1ddx[x]
Step 2.3.11
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
13limx→024x2+11
13limx→024x2+11
Step 2.4
Multiply the numerator by the reciprocal of the denominator.
13limx→024x2+1⋅1
Step 2.5
Multiply 24x2+1 by 1.
13limx→024x2+1
13limx→024x2+1
Step 3
Step 3.1
Move the term 2 outside of the limit because it is constant with respect to x.
13⋅2limx→014x2+1
Step 3.2
Split the limit using the Limits Quotient Rule on the limit as x approaches 0.
13⋅2limx→01limx→04x2+1
Step 3.3
Evaluate the limit of 1 which is constant as x approaches 0.
13⋅21limx→04x2+1
Step 3.4
Split the limit using the Sum of Limits Rule on the limit as x approaches 0.
13⋅21limx→04x2+limx→01
Step 3.5
Move the term 4 outside of the limit because it is constant with respect to x.
13⋅214limx→0x2+limx→01
Step 3.6
Move the exponent 2 from x2 outside the limit using the Limits Power Rule.
13⋅214(limx→0x)2+limx→01
Step 3.7
Evaluate the limit of 1 which is constant as x approaches 0.
13⋅214(limx→0x)2+1
13⋅214(limx→0x)2+1
Step 4
Evaluate the limit of x by plugging in 0 for x.
13⋅214⋅02+1
Step 5
Step 5.1
Combine 13 and 2.
23⋅14⋅02+1
Step 5.2
Simplify the denominator.
Step 5.2.1
Raising 0 to any positive power yields 0.
23⋅14⋅0+1
Step 5.2.2
Multiply 4 by 0.
23⋅10+1
Step 5.2.3
Add 0 and 1.
23⋅11
23⋅11
Step 5.3
Cancel the common factor of 1.
Step 5.3.1
Cancel the common factor.
23⋅11
Step 5.3.2
Rewrite the expression.
23⋅1
23⋅1
Step 5.4
Multiply 23 by 1.
23
23
Step 6
The result can be shown in multiple forms.
Exact Form:
23
Decimal Form:
0.‾6