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Calculus Examples
limx→0x+2x23ln(x+1)-3x
Step 1
Step 1.1
Take the limit of the numerator and the limit of the denominator.
limx→0x+2x2limx→03ln(x+1)-3x
Step 1.2
Evaluate the limit of the numerator.
Step 1.2.1
Split the limit using the Sum of Limits Rule on the limit as x approaches 0.
limx→0x+limx→02x2limx→03ln(x+1)-3x
Step 1.2.2
Move the term 2 outside of the limit because it is constant with respect to x.
limx→0x+2limx→0x2limx→03ln(x+1)-3x
Step 1.2.3
Move the exponent 2 from x2 outside the limit using the Limits Power Rule.
limx→0x+2(limx→0x)2limx→03ln(x+1)-3x
Step 1.2.4
Evaluate the limits by plugging in 0 for all occurrences of x.
Step 1.2.4.1
Evaluate the limit of x by plugging in 0 for x.
0+2(limx→0x)2limx→03ln(x+1)-3x
Step 1.2.4.2
Evaluate the limit of x by plugging in 0 for x.
0+2⋅02limx→03ln(x+1)-3x
0+2⋅02limx→03ln(x+1)-3x
Step 1.2.5
Simplify the answer.
Step 1.2.5.1
Simplify each term.
Step 1.2.5.1.1
Raising 0 to any positive power yields 0.
0+2⋅0limx→03ln(x+1)-3x
Step 1.2.5.1.2
Multiply 2 by 0.
0+0limx→03ln(x+1)-3x
0+0limx→03ln(x+1)-3x
Step 1.2.5.2
Add 0 and 0.
0limx→03ln(x+1)-3x
0limx→03ln(x+1)-3x
0limx→03ln(x+1)-3x
Step 1.3
Evaluate the limit of the denominator.
Step 1.3.1
Split the limit using the Sum of Limits Rule on the limit as x approaches 0.
0limx→03ln(x+1)-limx→03x
Step 1.3.2
Move the term 3 outside of the limit because it is constant with respect to x.
03limx→0ln(x+1)-limx→03x
Step 1.3.3
Move the limit inside the logarithm.
03ln(limx→0x+1)-limx→03x
Step 1.3.4
Split the limit using the Sum of Limits Rule on the limit as x approaches 0.
03ln(limx→0x+limx→01)-limx→03x
Step 1.3.5
Evaluate the limit of 1 which is constant as x approaches 0.
03ln(limx→0x+1)-limx→03x
Step 1.3.6
Move the term 3 outside of the limit because it is constant with respect to x.
03ln(limx→0x+1)-3limx→0x
Step 1.3.7
Evaluate the limits by plugging in 0 for all occurrences of x.
Step 1.3.7.1
Evaluate the limit of x by plugging in 0 for x.
03ln(0+1)-3limx→0x
Step 1.3.7.2
Evaluate the limit of x by plugging in 0 for x.
03ln(0+1)-3⋅0
03ln(0+1)-3⋅0
Step 1.3.8
Simplify the answer.
Step 1.3.8.1
Simplify each term.
Step 1.3.8.1.1
Add 0 and 1.
03ln(1)-3⋅0
Step 1.3.8.1.2
The natural logarithm of 1 is 0.
03⋅0-3⋅0
Step 1.3.8.1.3
Multiply 3 by 0.
00-3⋅0
Step 1.3.8.1.4
Multiply -3 by 0.
00+0
00+0
Step 1.3.8.2
Add 0 and 0.
00
Step 1.3.8.3
The expression contains a division by 0. The expression is undefined.
Undefined
00
Step 1.3.9
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.
limx→0x+2x23ln(x+1)-3x=limx→0ddx[x+2x2]ddx[3ln(x+1)-3x]
Step 3
Step 3.1
Differentiate the numerator and denominator.
limx→0ddx[x+2x2]ddx[3ln(x+1)-3x]
Step 3.2
By the Sum Rule, the derivative of x+2x2 with respect to x is ddx[x]+ddx[2x2].
limx→0ddx[x]+ddx[2x2]ddx[3ln(x+1)-3x]
Step 3.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→01+ddx[2x2]ddx[3ln(x+1)-3x]
Step 3.4
Evaluate ddx[2x2].
Step 3.4.1
Since 2 is constant with respect to x, the derivative of 2x2 with respect to x is 2ddx[x2].
limx→01+2ddx[x2]ddx[3ln(x+1)-3x]
Step 3.4.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
limx→01+2(2x)ddx[3ln(x+1)-3x]
Step 3.4.3
Multiply 2 by 2.
limx→01+4xddx[3ln(x+1)-3x]
limx→01+4xddx[3ln(x+1)-3x]
Step 3.5
Reorder terms.
limx→04x+1ddx[3ln(x+1)-3x]
Step 3.6
By the Sum Rule, the derivative of 3ln(x+1)-3x with respect to x is ddx[3ln(x+1)]+ddx[-3x].
limx→04x+1ddx[3ln(x+1)]+ddx[-3x]
Step 3.7
Evaluate ddx[3ln(x+1)].
Step 3.7.1
Since 3 is constant with respect to x, the derivative of 3ln(x+1) with respect to x is 3ddx[ln(x+1)].
limx→04x+13ddx[ln(x+1)]+ddx[-3x]
Step 3.7.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=ln(x) and g(x)=x+1.
Step 3.7.2.1
To apply the Chain Rule, set u as x+1.
limx→04x+13(ddu[ln(u)]ddx[x+1])+ddx[-3x]
Step 3.7.2.2
The derivative of ln(u) with respect to u is 1u.
limx→04x+13(1uddx[x+1])+ddx[-3x]
Step 3.7.2.3
Replace all occurrences of u with x+1.
limx→04x+13(1x+1ddx[x+1])+ddx[-3x]
limx→04x+13(1x+1ddx[x+1])+ddx[-3x]
Step 3.7.3
By the Sum Rule, the derivative of x+1 with respect to x is ddx[x]+ddx[1].
limx→04x+13(1x+1(ddx[x]+ddx[1]))+ddx[-3x]
Step 3.7.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→04x+13(1x+1(1+ddx[1]))+ddx[-3x]
Step 3.7.5
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
limx→04x+13(1x+1(1+0))+ddx[-3x]
Step 3.7.6
Add 1 and 0.
limx→04x+13(1x+1⋅1)+ddx[-3x]
Step 3.7.7
Multiply 1x+1 by 1.
limx→04x+131x+1+ddx[-3x]
Step 3.7.8
Combine 3 and 1x+1.
limx→04x+13x+1+ddx[-3x]
limx→04x+13x+1+ddx[-3x]
Step 3.8
Evaluate ddx[-3x].
Step 3.8.1
Since -3 is constant with respect to x, the derivative of -3x with respect to x is -3ddx[x].
limx→04x+13x+1-3ddx[x]
Step 3.8.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→04x+13x+1-3⋅1
Step 3.8.3
Multiply -3 by 1.
limx→04x+13x+1-3
limx→04x+13x+1-3
Step 3.9
Simplify.
Step 3.9.1
Combine terms.
Step 3.9.1.1
To write -3 as a fraction with a common denominator, multiply by x+1x+1.
limx→04x+13x+1-3⋅x+1x+1
Step 3.9.1.2
Combine -3 and x+1x+1.
limx→04x+13x+1+-3(x+1)x+1
Step 3.9.1.3
Combine the numerators over the common denominator.
limx→04x+13-3(x+1)x+1
limx→04x+13-3(x+1)x+1
Step 3.9.2
Simplify the numerator.
Step 3.9.2.1
Factor 3 out of 3-3(x+1).
Step 3.9.2.1.1
Factor 3 out of 3.
limx→04x+13(1)-3(x+1)x+1
Step 3.9.2.1.2
Factor 3 out of -3(x+1).
limx→04x+13(1)+3(-(x+1))x+1
Step 3.9.2.1.3
Factor 3 out of 3(1)+3(-(x+1)).
limx→04x+13(1-(x+1))x+1
limx→04x+13(1-(x+1))x+1
Step 3.9.2.2
Apply the distributive property.
limx→04x+13(1-x-1⋅1)x+1
Step 3.9.2.3
Multiply -1 by 1.
limx→04x+13(1-x-1)x+1
Step 3.9.2.4
Subtract 1 from 1.
limx→04x+13(-x+0)x+1
Step 3.9.2.5
Add -x and 0.
limx→04x+13⋅-1xx+1
Step 3.9.2.6
Combine exponents.
Step 3.9.2.6.1
Factor out negative.
limx→04x+1-(3x)x+1
Step 3.9.2.6.2
Multiply 3 by -1.
limx→04x+1-3xx+1
limx→04x+1-3xx+1
limx→04x+1-3xx+1
Step 3.9.3
Move the negative in front of the fraction.
limx→04x+1-3xx+1
limx→04x+1-3xx+1
limx→04x+1-3xx+1
Step 4
Multiply the numerator by the reciprocal of the denominator.
limx→0(4x+1)(-x+13x)
Step 5
Consider the left sided limit.
limx→0-(4x+1)(-x+13x)
Step 6
As the x values approach 0 from the left, the function values increase without bound.
∞
Step 7
Consider the right sided limit.
limx→0+(4x+1)(-x+13x)
Step 8
As the x values approach 0 from the right, the function values decrease without bound.
-∞
Step 9
Since the left sided and right sided limits are not equal, the limit does not exist.
Does not exist