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Calculus Examples
limx→-∞(x+2)(x+5)(x-5)13(x+3)(x+5)limx→−∞(x+2)(x+5)(x−5)13(x+3)(x+5)
Step 1
Step 1.1
Cancel the common factor.
limx→-∞(x+2)(x+5)(x-5)13(x+3)(x+5)
Step 1.2
Rewrite the expression.
limx→-∞(x+2)(x-5)13(x+3)
limx→-∞(x+2)(x-5)13(x+3)
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.
limx→-∞(x+2)(x-5)limx→-∞13(x+3)
Step 2.1.2
Evaluate the limit of the numerator.
Step 2.1.2.1
Apply the distributive property.
limx→-∞x(x-5)+2(x-5)limx→-∞13(x+3)
Step 2.1.2.2
Apply the distributive property.
limx→-∞x⋅x+x⋅-5+2(x-5)limx→-∞13(x+3)
Step 2.1.2.3
Apply the distributive property.
limx→-∞x⋅x+x⋅-5+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.4
Reorder x and -5.
limx→-∞x⋅x-5⋅x+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.5
Raise x to the power of 1.
limx→-∞x1x-5x+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.6
Raise x to the power of 1.
limx→-∞x1x1-5x+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.7
Use the power rule aman=am+n to combine exponents.
limx→-∞x1+1-5x+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.8
Simplify by adding terms.
Step 2.1.2.8.1
Add 1 and 1.
limx→-∞x2-5x+2x+2⋅-5limx→-∞13(x+3)
Step 2.1.2.8.2
Multiply 2 by -5.
limx→-∞x2-5x+2x-10limx→-∞13(x+3)
Step 2.1.2.8.3
Add -5x and 2x.
limx→-∞x2-3x-10limx→-∞13(x+3)
limx→-∞x2-3x-10limx→-∞13(x+3)
Step 2.1.2.9
The limit at negative infinity of a polynomial of even degree whose leading coefficient is positive is infinity.
∞limx→-∞13(x+3)
∞limx→-∞13(x+3)
Step 2.1.3
Evaluate the limit of the denominator.
Step 2.1.3.1
Simplify by multiplying through.
Step 2.1.3.1.1
Apply the distributive property.
∞limx→-∞13x+13⋅3
Step 2.1.3.1.2
Multiply 13 by 3.
∞limx→-∞13x+39
∞limx→-∞13x+39
Step 2.1.3.2
The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.
∞-∞
Step 2.1.3.3
Infinity divided by infinity is undefined.
Undefined
∞-∞
Step 2.1.4
Infinity divided by infinity is undefined.
Undefined
∞-∞
Step 2.2
Since ∞-∞ 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→-∞(x+2)(x-5)13(x+3)=limx→-∞ddx[(x+2)(x-5)]ddx[13(x+3)]
Step 2.3
Find the derivative of the numerator and denominator.
Step 2.3.1
Differentiate the numerator and denominator.
limx→-∞ddx[(x+2)(x-5)]ddx[13(x+3)]
Step 2.3.2
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x+2 and g(x)=x-5.
limx→-∞(x+2)ddx[x-5]+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.3
By the Sum Rule, the derivative of x-5 with respect to x is ddx[x]+ddx[-5].
limx→-∞(x+2)(ddx[x]+ddx[-5])+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→-∞(x+2)(1+ddx[-5])+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.5
Since -5 is constant with respect to x, the derivative of -5 with respect to x is 0.
limx→-∞(x+2)(1+0)+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.6
Add 1 and 0.
limx→-∞(x+2)⋅1+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.7
Multiply x+2 by 1.
limx→-∞x+2+(x-5)ddx[x+2]ddx[13(x+3)]
Step 2.3.8
By the Sum Rule, the derivative of x+2 with respect to x is ddx[x]+ddx[2].
limx→-∞x+2+(x-5)(ddx[x]+ddx[2])ddx[13(x+3)]
Step 2.3.9
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→-∞x+2+(x-5)(1+ddx[2])ddx[13(x+3)]
Step 2.3.10
Since 2 is constant with respect to x, the derivative of 2 with respect to x is 0.
limx→-∞x+2+(x-5)(1+0)ddx[13(x+3)]
Step 2.3.11
Add 1 and 0.
limx→-∞x+2+(x-5)⋅1ddx[13(x+3)]
Step 2.3.12
Multiply x-5 by 1.
limx→-∞x+2+x-5ddx[13(x+3)]
Step 2.3.13
Add x and x.
limx→-∞2x+2-5ddx[13(x+3)]
Step 2.3.14
Subtract 5 from 2.
limx→-∞2x-3ddx[13(x+3)]
Step 2.3.15
Since 13 is constant with respect to x, the derivative of 13(x+3) with respect to x is 13ddx[x+3].
limx→-∞2x-313ddx[x+3]
Step 2.3.16
By the Sum Rule, the derivative of x+3 with respect to x is ddx[x]+ddx[3].
limx→-∞2x-313(ddx[x]+ddx[3])
Step 2.3.17
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx→-∞2x-313(1+ddx[3])
Step 2.3.18
Since 3 is constant with respect to x, the derivative of 3 with respect to x is 0.
limx→-∞2x-313(1+0)
Step 2.3.19
Add 1 and 0.
limx→-∞2x-313⋅1
Step 2.3.20
Multiply 13 by 1.
limx→-∞2x-313
limx→-∞2x-313
limx→-∞2x-313
Step 3
Split the fraction 2x-313 into two fractions.
limx→-∞2x13+-313
Step 4
The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.
-∞