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Calculus Examples
limx→0xarctan(2x)limx→0xarctan(2x)
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→0xlimx→0arctan(2x)limx→0xlimx→0arctan(2x)
Step 1.1.2
Evaluate the limit of xx by plugging in 00 for xx.
0limx→0arctan(2x)0limx→0arctan(2x)
Step 1.1.3
Evaluate the limit of the denominator.
Step 1.1.3.1
Evaluate the limit of xx by plugging in 00 for xx.
0000
Step 1.1.3.2
Substitute tt for 2x2x and let tt approach 00 since limx→02x=0limx→02x=0.
0limt→0arctan(t)0limt→0arctan(t)
Step 1.1.3.3
Evaluate the limits by plugging in 00 for all occurrences of tt.
Step 1.1.3.3.1
Evaluate the limit of xx by plugging in 00 for xx.
0arctan(0)0arctan(0)
Step 1.1.3.3.2
The exact value of arctan(0)arctan(0) is 00.
0000
Step 1.1.3.3.3
The expression contains a division by 00. The expression is undefined.
Undefined
0000
Step 1.1.3.4
The expression contains a division by 00. The expression is undefined.
Undefined
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→0xarctan(2x)=limx→0ddx[x]ddx[arctan(2x)]limx→0xarctan(2x)=limx→0ddx[x]ddx[arctan(2x)]
Step 1.3
Find the derivative of the numerator and denominator.
Step 1.3.1
Differentiate the numerator and denominator.
limx→0ddx[x]ddx[arctan(2x)]limx→0ddx[x]ddx[arctan(2x)]
Step 1.3.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
limx→01ddx[arctan(2x)]limx→01ddx[arctan(2x)]
Step 1.3.3
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)=arctan(x) and g(x)=2x.
Step 1.3.3.1
To apply the Chain Rule, set u as 2x.
limx→01ddu[arctan(u)]ddx[2x]
Step 1.3.3.2
The derivative of arctan(u) with respect to u is 11+u2.
limx→0111+u2ddx[2x]
Step 1.3.3.3
Replace all occurrences of u with 2x.
limx→0111+(2x)2ddx[2x]
limx→0111+(2x)2ddx[2x]
Step 1.3.4
Factor 2 out of 2x.
limx→0111+(2(x))2ddx[2x]
Step 1.3.5
Apply the product rule to 2(x).
limx→0111+22x2ddx[2x]
Step 1.3.6
Raise 2 to the power of 2.
limx→0111+4x2ddx[2x]
Step 1.3.7
Since 2 is constant with respect to x, the derivative of 2x with respect to x is 2ddx[x].
limx→0111+4x2(2ddx[x])
Step 1.3.8
Combine 2 and 11+4x2.
limx→0121+4x2ddx[x]
Step 1.3.9
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→0121+4x2⋅1
Step 1.3.10
Multiply 21+4x2 by 1.
limx→0121+4x2
Step 1.3.11
Reorder terms.
limx→0124x2+1
limx→0124x2+1
Step 1.4
Multiply the numerator by the reciprocal of the denominator.
limx→014x2+12
Step 1.5
Multiply 4x2+12 by 1.
limx→04x2+12
limx→04x2+12
Step 2
Step 2.1
Move the term 12 outside of the limit because it is constant with respect to x.
12limx→04x2+1
Step 2.2
Split the limit using the Sum of Limits Rule on the limit as x approaches 0.
12(limx→04x2+limx→01)
Step 2.3
Move the term 4 outside of the limit because it is constant with respect to x.
12(4limx→0x2+limx→01)
Step 2.4
Move the exponent 2 from x2 outside the limit using the Limits Power Rule.
12(4(limx→0x)2+limx→01)
Step 2.5
Evaluate the limit of 1 which is constant as x approaches 0.
12(4(limx→0x)2+1)
12(4(limx→0x)2+1)
Step 3
Evaluate the limit of x by plugging in 0 for x.
12(4⋅02+1)
Step 4
Step 4.1
Simplify each term.
Step 4.1.1
Raising 0 to any positive power yields 0.
12(4⋅0+1)
Step 4.1.2
Multiply 4 by 0.
12(0+1)
12(0+1)
Step 4.2
Add 0 and 1.
12⋅1
Step 4.3
Multiply 12 by 1.
12
12
Step 5
The result can be shown in multiple forms.
Exact Form:
12
Decimal Form:
0.5