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Calculus Examples
limx→81√x-9x-81limx→81√x−9x−81
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→81√x-9limx→81x-81limx→81√x−9limx→81x−81
Step 1.1.2
Evaluate the limit of the numerator.
Step 1.1.2.1
Evaluate the limit.
Step 1.1.2.1.1
Split the limit using the Sum of Limits Rule on the limit as xx approaches 8181.
limx→81√x-limx→819limx→81x-81limx→81√x−limx→819limx→81x−81
Step 1.1.2.1.2
Move the limit under the radical sign.
√limx→81x-limx→819limx→81x-81√limx→81x−limx→819limx→81x−81
Step 1.1.2.1.3
Evaluate the limit of 99 which is constant as xx approaches 8181.
√limx→81x-1⋅9limx→81x-81√limx→81x−1⋅9limx→81x−81
√limx→81x-1⋅9limx→81x-81√limx→81x−1⋅9limx→81x−81
Step 1.1.2.2
Evaluate the limit of xx by plugging in 8181 for xx.
√81-1⋅9limx→81x-81√81−1⋅9limx→81x−81
Step 1.1.2.3
Simplify the answer.
Step 1.1.2.3.1
Simplify each term.
Step 1.1.2.3.1.1
Rewrite 8181 as 9292.
√92-1⋅9limx→81x-81√92−1⋅9limx→81x−81
Step 1.1.2.3.1.2
Pull terms out from under the radical, assuming positive real numbers.
9-1⋅9limx→81x-819−1⋅9limx→81x−81
Step 1.1.2.3.1.3
Multiply -1−1 by 99.
9-9limx→81x-819−9limx→81x−81
9-9limx→81x-819−9limx→81x−81
Step 1.1.2.3.2
Subtract 99 from 99.
0limx→81x-810limx→81x−81
0limx→81x-810limx→81x−81
0limx→81x-810limx→81x−81
Step 1.1.3
Evaluate the limit of the denominator.
Step 1.1.3.1
Evaluate the limit.
Step 1.1.3.1.1
Split the limit using the Sum of Limits Rule on the limit as xx approaches 8181.
0limx→81x-limx→81810limx→81x−limx→8181
Step 1.1.3.1.2
Evaluate the limit of 8181 which is constant as xx approaches 8181.
0limx→81x-1⋅810limx→81x−1⋅81
0limx→81x-1⋅810limx→81x−1⋅81
Step 1.1.3.2
Evaluate the limit of xx by plugging in 8181 for xx.
081-1⋅81081−1⋅81
Step 1.1.3.3
Simplify the answer.
Step 1.1.3.3.1
Multiply -1−1 by 8181.
081-81081−81
Step 1.1.3.3.2
Subtract 8181 from 8181.
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→81√x-9x-81=limx→81ddx[√x-9]ddx[x-81]limx→81√x−9x−81=limx→81ddx[√x−9]ddx[x−81]
Step 1.3
Find the derivative of the numerator and denominator.
Step 1.3.1
Differentiate the numerator and denominator.
limx→81ddx[√x-9]ddx[x-81]limx→81ddx[√x−9]ddx[x−81]
Step 1.3.2
By the Sum Rule, the derivative of √x-9√x−9 with respect to xx is ddx[√x]+ddx[-9]ddx[√x]+ddx[−9].
limx→81ddx[√x]+ddx[-9]ddx[x-81]limx→81ddx[√x]+ddx[−9]ddx[x−81]
Step 1.3.3
Evaluate ddx[√x]ddx[√x].
Step 1.3.3.1
Use n√ax=axnn√ax=axn to rewrite √x√x as x12x12.
limx→81ddx[x12]+ddx[-9]ddx[x-81]limx→81ddx[x12]+ddx[−9]ddx[x−81]
Step 1.3.3.2
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=12n=12.
limx→8112x12-1+ddx[-9]ddx[x-81]limx→8112x12−1+ddx[−9]ddx[x−81]
Step 1.3.3.3
To write -1−1 as a fraction with a common denominator, multiply by 2222.
limx→8112x12-1⋅22+ddx[-9]ddx[x-81]limx→8112x12−1⋅22+ddx[−9]ddx[x−81]
Step 1.3.3.4
Combine -1−1 and 2222.
limx→8112x12+-1⋅22+ddx[-9]ddx[x-81]limx→8112x12+−1⋅22+ddx[−9]ddx[x−81]
Step 1.3.3.5
Combine the numerators over the common denominator.
limx→8112x1-1⋅22+ddx[-9]ddx[x-81]limx→8112x1−1⋅22+ddx[−9]ddx[x−81]
Step 1.3.3.6
Simplify the numerator.
Step 1.3.3.6.1
Multiply -1−1 by 22.
limx→8112x1-22+ddx[-9]ddx[x-81]limx→8112x1−22+ddx[−9]ddx[x−81]
Step 1.3.3.6.2
Subtract 22 from 11.
limx→8112x-12+ddx[-9]ddx[x-81]limx→8112x−12+ddx[−9]ddx[x−81]
limx→8112x-12+ddx[-9]ddx[x-81]limx→8112x−12+ddx[−9]ddx[x−81]
Step 1.3.3.7
Move the negative in front of the fraction.
limx→8112x-12+ddx[-9]ddx[x-81]limx→8112x−12+ddx[−9]ddx[x−81]
limx→8112x-12+ddx[-9]ddx[x-81]limx→8112x−12+ddx[−9]ddx[x−81]
Step 1.3.4
Since -9−9 is constant with respect to xx, the derivative of -9−9 with respect to xx is 00.
limx→8112x-12+0ddx[x-81]limx→8112x−12+0ddx[x−81]
Step 1.3.5
Simplify.
Step 1.3.5.1
Rewrite the expression using the negative exponent rule b-n=1bnb−n=1bn.
limx→8112⋅1x12+0ddx[x-81]limx→8112⋅1x12+0ddx[x−81]
Step 1.3.5.2
Combine terms.
Step 1.3.5.2.1
Multiply 1212 by 1x121x12.
limx→8112x12+0ddx[x-81]limx→8112x12+0ddx[x−81]
Step 1.3.5.2.2
Add 12x1212x12 and 00.
limx→8112x12ddx[x-81]limx→8112x12ddx[x−81]
limx→8112x12ddx[x-81]limx→8112x12ddx[x−81]
limx→8112x12ddx[x-81]limx→8112x12ddx[x−81]
Step 1.3.6
By the Sum Rule, the derivative of x-81x−81 with respect to xx is ddx[x]+ddx[-81]ddx[x]+ddx[−81].
limx→8112x12ddx[x]+ddx[-81]limx→8112x12ddx[x]+ddx[−81]
Step 1.3.7
Differentiate using the Power Rule which states that ddx[xn]ddx[xn] is nxn-1nxn−1 where n=1n=1.
limx→8112x121+ddx[-81]limx→8112x121+ddx[−81]
Step 1.3.8
Since -81−81 is constant with respect to xx, the derivative of -81−81 with respect to xx is 00.
limx→8112x121+0limx→8112x121+0
Step 1.3.9
Add 11 and 00.
limx→8112x121limx→8112x121
limx→8112x121
Step 1.4
Multiply the numerator by the reciprocal of the denominator.
limx→8112x12⋅1
Step 1.5
Rewrite x12 as √x.
limx→8112√x⋅1
Step 1.6
Multiply 12√x by 1.
limx→8112√x
limx→8112√x
Step 2
Step 2.1
Move the term 12 outside of the limit because it is constant with respect to x.
12limx→811√x
Step 2.2
Split the limit using the Limits Quotient Rule on the limit as x approaches 81.
12⋅limx→811limx→81√x
Step 2.3
Evaluate the limit of 1 which is constant as x approaches 81.
12⋅1limx→81√x
Step 2.4
Move the limit under the radical sign.
12⋅1√limx→81x
12⋅1√limx→81x
Step 3
Evaluate the limit of x by plugging in 81 for x.
12⋅1√81
Step 4
Step 4.1
Simplify the denominator.
Step 4.1.1
Rewrite 81 as 92.
12⋅1√92
Step 4.1.2
Pull terms out from under the radical, assuming positive real numbers.
12⋅19
12⋅19
Step 4.2
Multiply 12⋅19.
Step 4.2.1
Multiply 12 by 19.
12⋅9
Step 4.2.2
Multiply 2 by 9.
118
118
118
Step 5
The result can be shown in multiple forms.
Exact Form:
118
Decimal Form:
0.0‾5