Calculus Examples

$lim_{x \to - b} \frac{{(x + b)}^{7} + {(x + b)}^{10}}{4 (x + b)}$

Step 1

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{7} + {(x + b)}^{10}}{x + b}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 2.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1}{4} \cdot \frac{lim_{x \to - b} {(x + b)}^{7} + {(x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{lim_{x \to - b} {(x + b)}^{7} + lim_{x \to - b} {(x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.2

Move the exponent $7$ from ${(x + b)}^{7}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + b)}^{7} + lim_{x \to - b} {(x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + lim_{x \to - b} b)}^{7} + lim_{x \to - b} {(x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.4

Evaluate the limit of $b$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + b)}^{7} + lim_{x \to - b} {(x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.5

Move the exponent $10$ from ${(x + b)}^{10}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + b)}^{7} + {(lim_{x \to - b} x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + b)}^{7} + {(lim_{x \to - b} x + lim_{x \to - b} b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.7

Evaluate the limit of $b$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{{(lim_{x \to - b} x + b)}^{7} + {(lim_{x \to - b} x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.8

Evaluate the limits by plugging in $- b$ for all occurrences of $x$ .

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Step 2.1.2.8.1

Evaluate the limit of $x$ by plugging in $- b$ for $x$ .

$\frac{1}{4} \cdot \frac{{(- b + b)}^{7} + {(lim_{x \to - b} x + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.8.2

Evaluate the limit of $x$ by plugging in $- b$ for $x$ .

$\frac{1}{4} \cdot \frac{{(- b + b)}^{7} + {(- b + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.9

Simplify the answer.

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Step 2.1.2.9.1

Combine the opposite terms in ${(- b + b)}^{7} + {(- b + b)}^{10}$ .

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Step 2.1.2.9.1.1

Add $- b$ and $b$ .

$\frac{1}{4} \cdot \frac{0^{7} + {(- b + b)}^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.9.1.2

Add $- b$ and $b$ .

$\frac{1}{4} \cdot \frac{0^{7} + 0^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.9.2

Simplify each term.

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Step 2.1.2.9.2.1

Raising $0$ to any positive power yields $0$ .

$\frac{1}{4} \cdot \frac{0 + 0^{10}}{lim_{x \to - b} x + b}$

Step 2.1.2.9.2.2

Raising $0$ to any positive power yields $0$ .

$\frac{1}{4} \cdot \frac{0 + 0}{lim_{x \to - b} x + b}$

Step 2.1.2.9.3

Add $0$ and $0$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - b} x + b}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Evaluate the limit.

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Step 2.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - b} x + lim_{x \to - b} b}$

Step 2.1.3.1.2

Evaluate the limit of $b$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} \cdot \frac{0}{lim_{x \to - b} x + b}$

Step 2.1.3.2

Evaluate the limit of $x$ by plugging in $- b$ for $x$ .

$\frac{1}{4} \cdot \frac{0}{- b + b}$

Step 2.1.3.3

Add $- b$ and $b$ .

$\frac{1}{4} \cdot \frac{0}{0}$

Step 2.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{4} \cdot \frac{0}{0}$

Step 2.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{4} \cdot \frac{0}{0}$

Step 2.2

Since $\frac{0}{0}$ 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.

$lim_{x \to - b} \frac{{(x + b)}^{7} + {(x + b)}^{10}}{x + b} = lim_{x \to - b} \frac{\frac{d}{d x} [{(x + b)}^{7} + {(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{4} lim_{x \to - b} \frac{\frac{d}{d x} [{(x + b)}^{7} + {(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.2

By the Sum Rule, the derivative of ${(x + b)}^{7} + {(x + b)}^{10}$ with respect to $x$ is $\frac{d}{d x} [{(x + b)}^{7}] + \frac{d}{d x} [{(x + b)}^{10}]$ .

$\frac{1}{4} lim_{x \to - b} \frac{\frac{d}{d x} [{(x + b)}^{7}] + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3

Evaluate $\frac{d}{d x} [{(x + b)}^{7}]$ .

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Step 2.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{7}$ and $g (x) = x + b$ .

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Step 2.3.3.1.1

To apply the Chain Rule, set $u_{1}$ as $x + b$ .

$\frac{1}{4} lim_{x \to - b} \frac{\frac{d}{d u_{1}} [{u_{1}}^{7}] \frac{d}{d x} [x + b] + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 7$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {u_{1}}^{6} \frac{d}{d x} [x + b] + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.1.3

Replace all occurrences of $u_{1}$ with $x + b$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} \frac{d}{d x} [x + b] + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.2

By the Sum Rule, the derivative of $x + b$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [b]$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} (\frac{d}{d x} [x] + \frac{d}{d x} [b]) + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} (1 + \frac{d}{d x} [b]) + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.4

Since $b$ is constant with respect to $x$ , the derivative of $b$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} (1 + 0) + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.5

Add $1$ and $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} \cdot 1 + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.3.6

Multiply $7$ by $1$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + \frac{d}{d x} [{(x + b)}^{10}]}{\frac{d}{d x} [x + b]}$

Step 2.3.4

Evaluate $\frac{d}{d x} [{(x + b)}^{10}]$ .

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Step 2.3.4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{10}$ and $g (x) = x + b$ .

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Step 2.3.4.1.1

To apply the Chain Rule, set $u_{2}$ as $x + b$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + \frac{d}{d u_{2}} [{u_{2}}^{10}] \frac{d}{d x} [x + b]}{\frac{d}{d x} [x + b]}$

Step 2.3.4.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 10$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {u_{2}}^{9} \frac{d}{d x} [x + b]}{\frac{d}{d x} [x + b]}$

Step 2.3.4.1.3

Replace all occurrences of $u_{2}$ with $x + b$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9} \frac{d}{d x} [x + b]}{\frac{d}{d x} [x + b]}$

Step 2.3.4.2

By the Sum Rule, the derivative of $x + b$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [b]$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9} (\frac{d}{d x} [x] + \frac{d}{d x} [b])}{\frac{d}{d x} [x + b]}$

Step 2.3.4.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9} (1 + \frac{d}{d x} [b])}{\frac{d}{d x} [x + b]}$

Step 2.3.4.4

Since $b$ is constant with respect to $x$ , the derivative of $b$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9} (1 + 0)}{\frac{d}{d x} [x + b]}$

Step 2.3.4.5

Add $1$ and $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9} \cdot 1}{\frac{d}{d x} [x + b]}$

Step 2.3.4.6

Multiply $10$ by $1$ .

$\frac{1}{4} lim_{x \to - b} \frac{7 {(x + b)}^{6} + 10 {(x + b)}^{9}}{\frac{d}{d x} [x + b]}$

Step 2.3.5

Factor ${(x + b)}^{6}$ out of $7 {(x + b)}^{6} + 10 {(x + b)}^{9}$ .

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Step 2.3.5.1

Factor ${(x + b)}^{6}$ out of $7 {(x + b)}^{6}$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} \cdot 7 + 10 {(x + b)}^{9}}{\frac{d}{d x} [x + b]}$

Step 2.3.5.2

Factor ${(x + b)}^{6}$ out of $10 {(x + b)}^{9}$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} \cdot 7 + {(x + b)}^{6} \cdot 10 {(x + b)}^{3}}{\frac{d}{d x} [x + b]}$

Step 2.3.5.3

Factor ${(x + b)}^{6}$ out of ${(x + b)}^{6} \cdot 7 + {(x + b)}^{6} (10 {(x + b)}^{3})$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} (7 + 10 {(x + b)}^{3})}{\frac{d}{d x} [x + b]}$

Step 2.3.6

By the Sum Rule, the derivative of $x + b$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [b]$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} (7 + 10 {(x + b)}^{3})}{\frac{d}{d x} [x] + \frac{d}{d x} [b]}$

Step 2.3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} (7 + 10 {(x + b)}^{3})}{1 + \frac{d}{d x} [b]}$

Step 2.3.8

Since $b$ is constant with respect to $x$ , the derivative of $b$ with respect to $x$ is $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} (7 + 10 {(x + b)}^{3})}{1 + 0}$

Step 2.3.9

Add $1$ and $0$ .

$\frac{1}{4} lim_{x \to - b} \frac{{(x + b)}^{6} (7 + 10 {(x + b)}^{3})}{1}$

Step 2.4

Divide ${(x + b)}^{6} (7 + 10 {(x + b)}^{3})$ by $1$ .

$\frac{1}{4} lim_{x \to - b} {(x + b)}^{6} (7 + 10 {(x + b)}^{3})$

Step 3

Evaluate the limit.

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Step 3.1

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} lim_{x \to - b} {(x + b)}^{6} \cdot lim_{x \to - b} 7 + 10 {(x + b)}^{3}$

Step 3.2

Move the exponent $6$ from ${(x + b)}^{6}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot lim_{x \to - b} 7 + 10 {(x + b)}^{3}$

Step 3.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + lim_{x \to - b} b)}^{6} \cdot lim_{x \to - b} 7 + 10 {(x + b)}^{3}$

Step 3.4

Evaluate the limit of $b$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot lim_{x \to - b} 7 + 10 {(x + b)}^{3}$

Step 3.5

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (lim_{x \to - b} 7 + lim_{x \to - b} 10 {(x + b)}^{3})$

Step 3.6

Evaluate the limit of $7$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (7 + lim_{x \to - b} 10 {(x + b)}^{3})$

Step 3.7

Move the term $10$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (7 + 10 lim_{x \to - b} {(x + b)}^{3})$

Step 3.8

Move the exponent $3$ from ${(x + b)}^{3}$ outside the limit using the Limits Power Rule.

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (7 + 10 {(lim_{x \to - b} x + b)}^{3})$

Step 3.9

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (7 + 10 {(lim_{x \to - b} x + lim_{x \to - b} b)}^{3})$

Step 3.10

Evaluate the limit of $b$ which is constant as $x$ approaches $- b$ .

$\frac{1}{4} {(lim_{x \to - b} x + b)}^{6} \cdot (7 + 10 {(lim_{x \to - b} x + b)}^{3})$

Step 4

Evaluate the limits by plugging in $- b$ for all occurrences of $x$ .

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Step 4.1

Evaluate the limit of $x$ by plugging in $- b$ for $x$ .

$\frac{1}{4} {(- b + b)}^{6} \cdot (7 + 10 {(lim_{x \to - b} x + b)}^{3})$

Step 4.2

Evaluate the limit of $x$ by plugging in $- b$ for $x$ .

$\frac{1}{4} {(- b + b)}^{6} \cdot (7 + 10 {(- b + b)}^{3})$

Step 5

Simplify the answer.

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Step 5.1

Add $- b$ and $b$ .

$\frac{1}{4} \cdot 0^{6} \cdot (7 + 10 {(- b + b)}^{3})$

Step 5.2

Raising $0$ to any positive power yields $0$ .

$\frac{1}{4} \cdot 0 \cdot (7 + 10 {(- b + b)}^{3})$

Step 5.3

Multiply $\frac{1}{4}$ by $0$ .

$0 \cdot (7 + 10 {(- b + b)}^{3})$

Step 5.4

Add $- b$ and $b$ .

$0 \cdot (7 + 10 \cdot 0^{3})$

Step 5.5

Simplify each term.

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Step 5.5.1

Raising $0$ to any positive power yields $0$ .

$0 \cdot (7 + 10 \cdot 0)$

Step 5.5.2

Multiply $10$ by $0$ .

$0 \cdot (7 + 0)$

Step 5.6

Add $7$ and $0$ .

$0 \cdot 7$

Step 5.7

Multiply $0$ by $7$ .

$0$