Calculus Examples

$lim_{x \to - 3} \frac{4 \ln (2 x + 7)}{2 \tan (- 6 - 2 x)}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to - 3} 4 \ln (2 x + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 lim_{x \to - 3} \ln (2 x + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.1.2

Move the limit inside the logarithm.

$\frac{4 \ln (lim_{x \to - 3} 2 x + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{4 \ln (lim_{x \to - 3} 2 x + lim_{x \to - 3} 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.1.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 \ln (2 lim_{x \to - 3} x + lim_{x \to - 3} 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.1.5

Evaluate the limit of $7$ which is constant as $x$ approaches $- 3$ .

$\frac{4 \ln (2 lim_{x \to - 3} x + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{4 \ln (2 \cdot - 3 + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Multiply $2$ by $- 3$ .

$\frac{4 \ln (- 6 + 7)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.3.2

Add $- 6$ and $7$ .

$\frac{4 \ln (1)}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.3.3

The natural logarithm of $1$ is $0$ .

$\frac{4 \cdot 0}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.2.3.4

Multiply $4$ by $0$ .

$\frac{0}{lim_{x \to - 3} 2 \tan (- 6 - 2 x)}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{2 lim_{x \to - 3} \tan (- 6 - 2 x)}$

Step 1.3.1.2

Move the limit inside the trig function because tangent is continuous.

$\frac{0}{2 \tan (lim_{x \to - 3} - 6 - 2 x)}$

Step 1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{0}{2 \tan (- lim_{x \to - 3} 6 - lim_{x \to - 3} 2 x)}$

Step 1.3.1.4

Evaluate the limit of $6$ which is constant as $x$ approaches $- 3$ .

$\frac{0}{2 \tan (- 1 \cdot 6 - lim_{x \to - 3} 2 x)}$

Step 1.3.1.5

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{2 \tan (- 1 \cdot 6 - 1 \cdot 2 lim_{x \to - 3} x)}$

Step 1.3.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{0}{2 \tan (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Multiply $- 1$ by $6$ .

$\frac{0}{2 \tan (- 6 - 1 \cdot 2 \cdot - 3)}$

Step 1.3.3.1.2

Multiply $- 1 \cdot 2 \cdot - 3$ .

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Step 1.3.3.1.2.1

Multiply $- 1$ by $2$ .

$\frac{0}{2 \tan (- 6 - 2 \cdot - 3)}$

Step 1.3.3.1.2.2

Multiply $- 2$ by $- 3$ .

$\frac{0}{2 \tan (- 6 + 6)}$

Step 1.3.3.2

Add $- 6$ and $6$ .

$\frac{0}{2 \tan (0)}$

Step 1.3.3.3

The exact value of $\tan (0)$ is $0$ .

$\frac{0}{2 \cdot 0}$

Step 1.3.3.4

Multiply $2$ by $0$ .

$\frac{0}{0}$

Step 1.3.3.5

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{x \to - 3} \frac{4 \ln (2 x + 7)}{2 \tan (- 6 - 2 x)} = lim_{x \to - 3} \frac{\frac{d}{d x} [4 \ln (2 x + 7)]}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - 3} \frac{\frac{d}{d x} [4 \ln (2 x + 7)]}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 \ln (2 x + 7)$ with respect to $x$ is $4 \frac{d}{d x} [\ln (2 x + 7)]$ .

$lim_{x \to - 3} \frac{4 \frac{d}{d x} [\ln (2 x + 7)]}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x + 7$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $2 x + 7$ .

$lim_{x \to - 3} \frac{4 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x + 7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.3.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$lim_{x \to - 3} \frac{4 (\frac{1}{u} \frac{d}{d x} [2 x + 7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.3.3

Replace all occurrences of $u$ with $2 x + 7$ .

$lim_{x \to - 3} \frac{4 (\frac{1}{2 x + 7} \frac{d}{d x} [2 x + 7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.4

Combine $\frac{1}{2 x + 7}$ and $4$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} \frac{d}{d x} [2 x + 7]}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.5

By the Sum Rule, the derivative of $2 x + 7$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [7]$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} (\frac{d}{d x} [2 x] + \frac{d}{d x} [7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.6

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} (2 \frac{d}{d x} [x] + \frac{d}{d x} [7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.7

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

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} (2 \cdot 1 + \frac{d}{d x} [7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.8

Multiply $2$ by $1$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} (2 + \frac{d}{d x} [7])}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.9

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} (2 + 0)}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.10

Add $2$ and $0$ .

$lim_{x \to - 3} \frac{\frac{4}{2 x + 7} \cdot 2}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.11

Combine $\frac{4}{2 x + 7}$ and $2$ .

$lim_{x \to - 3} \frac{\frac{4 \cdot 2}{2 x + 7}}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.12

Multiply $4$ by $2$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{\frac{d}{d x} [2 \tan (- 6 - 2 x)]}$

Step 3.13

Since $2$ is constant with respect to $x$ , the derivative of $2 \tan (- 6 - 2 x)$ with respect to $x$ is $2 \frac{d}{d x} [\tan (- 6 - 2 x)]$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \frac{d}{d x} [\tan (- 6 - 2 x)]}$

Step 3.14

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = - 6 - 2 x$ .

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Step 3.14.1

To apply the Chain Rule, set $u$ as $- 6 - 2 x$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 (\frac{d}{d u} [\tan (u)] \frac{d}{d x} [- 6 - 2 x])}$

Step 3.14.2

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 (\sec^{2} (u) \frac{d}{d x} [- 6 - 2 x])}$

Step 3.14.3

Replace all occurrences of $u$ with $- 6 - 2 x$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 (\sec^{2} (- 6 - 2 x) \frac{d}{d x} [- 6 - 2 x])}$

Step 3.15

Remove parentheses.

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \sec^{2} (- 6 - 2 x) \frac{d}{d x} [- 6 - 2 x]}$

Step 3.16

By the Sum Rule, the derivative of $- 6 - 2 x$ with respect to $x$ is $\frac{d}{d x} [- 6] + \frac{d}{d x} [- 2 x]$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \sec^{2} (- 6 - 2 x) (\frac{d}{d x} [- 6] + \frac{d}{d x} [- 2 x])}$

Step 3.17

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \sec^{2} (- 6 - 2 x) (0 + \frac{d}{d x} [- 2 x])}$

Step 3.18

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \sec^{2} (- 6 - 2 x) \frac{d}{d x} [- 2 x]}$

Step 3.19

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{2 \sec^{2} (- 6 - 2 x) (- 2 \frac{d}{d x} [x])}$

Step 3.20

Multiply $- 2$ by $2$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{- 4 \sec^{2} (- 6 - 2 x) \frac{d}{d x} [x]}$

Step 3.21

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

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{- 4 \sec^{2} (- 6 - 2 x) \cdot 1}$

Step 3.22

Multiply $- 4$ by $1$ .

$lim_{x \to - 3} \frac{\frac{8}{2 x + 7}}{- 4 \sec^{2} (- 6 - 2 x)}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 3} \frac{8}{2 x + 7} \cdot \frac{1}{- 4 \sec^{2} (- 6 - 2 x)}$

Step 5

Simplify terms.

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Step 5.1

Multiply $\frac{8}{2 x + 7}$ by $\frac{1}{- 4 \sec^{2} (- 6 - 2 x)}$ .

$lim_{x \to - 3} \frac{8}{(2 x + 7) (- 4 \sec^{2} (- 6 - 2 x))}$

Step 5.2

Cancel the common factor of $8$ and $- 4$ .

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Step 5.2.1

Factor $4$ out of $8$ .

$lim_{x \to - 3} \frac{4 (2)}{(2 x + 7) (- 4 \sec^{2} (- 6 - 2 x))}$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $4$ out of $(2 x + 7) (- 4 \sec^{2} (- 6 - 2 x))$ .

$lim_{x \to - 3} \frac{4 (2)}{4 ((2 x + 7) (- \sec^{2} (- 6 - 2 x)))}$

Step 5.2.2.2

Cancel the common factor.

$lim_{x \to - 3} \frac{4 \cdot 2}{4 ((2 x + 7) (- \sec^{2} (- 6 - 2 x)))}$

Step 5.2.2.3

Rewrite the expression.

$lim_{x \to - 3} \frac{2}{(2 x + 7) (- \sec^{2} (- 6 - 2 x))}$

Step 6

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to - 3} \frac{1}{(2 x + 7) (- \sec^{2} (- 6 - 2 x))}$

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 3$ .

$2 \frac{lim_{x \to - 3} 1}{lim_{x \to - 3} (2 x + 7) \cdot - 1 \sec^{2} (- 6 - 2 x)}$

Step 8

Evaluate the limit of $1$ which is constant as $x$ approaches $- 3$ .

$2 \frac{1}{lim_{x \to - 3} (2 x + 7) \cdot - 1 \sec^{2} (- 6 - 2 x)}$

Step 9

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{1}{- lim_{x \to - 3} (2 x + 7) \sec^{2} (- 6 - 2 x)}$

Step 10

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- 3$ .

$2 \frac{1}{- lim_{x \to - 3} 2 x + 7 \cdot lim_{x \to - 3} \sec^{2} (- 6 - 2 x)}$

Step 11

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$2 \frac{1}{- (lim_{x \to - 3} 2 x + lim_{x \to - 3} 7) \cdot lim_{x \to - 3} \sec^{2} (- 6 - 2 x)}$

Step 12

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{1}{- (2 lim_{x \to - 3} x + lim_{x \to - 3} 7) \cdot lim_{x \to - 3} \sec^{2} (- 6 - 2 x)}$

Step 13

Evaluate the limit of $7$ which is constant as $x$ approaches $- 3$ .

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot lim_{x \to - 3} \sec^{2} (- 6 - 2 x)}$

Step 14

Move the exponent $2$ from $\sec^{2} (- 6 - 2 x)$ outside the limit using the Limits Power Rule.

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot {(lim_{x \to - 3} \sec (- 6 - 2 x))}^{2}}$

Step 15

Move the limit inside the trig function because secant is continuous.

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot \sec^{2} (lim_{x \to - 3} - 6 - 2 x)}$

Step 16

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot \sec^{2} (- lim_{x \to - 3} 6 - lim_{x \to - 3} 2 x)}$

Step 17

Evaluate the limit of $6$ which is constant as $x$ approaches $- 3$ .

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot \sec^{2} (- 1 \cdot 6 - lim_{x \to - 3} 2 x)}$

Step 18

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{1}{- (2 lim_{x \to - 3} x + 7) \cdot \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 lim_{x \to - 3} x)}$

Step 19

Evaluate the limits by plugging in $- 3$ for all occurrences of $x$ .

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Step 19.1

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$2 \frac{1}{- (2 \cdot - 3 + 7) \cdot \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 lim_{x \to - 3} x)}$

Step 19.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$2 \frac{1}{- (2 \cdot - 3 + 7) \cdot \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)}$

Step 20

Simplify the answer.

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Step 20.1

Cancel the common factor of $1$ and $- 1$ .

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Step 20.1.1

Rewrite $1$ as $- 1 (- 1)$ .

$2 \frac{- 1 (- 1)}{- (2 \cdot - 3 + 7) \cdot \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)}$

Step 20.1.2

Move the negative in front of the fraction.

$2 (- \frac{1}{(2 \cdot - 3 + 7) \cdot \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)})$

Step 20.2

Simplify the denominator.

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Step 20.2.1

Multiply $2$ by $- 3$ .

$2 (- \frac{1}{(- 6 + 7) \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)})$

Step 20.2.2

Add $- 6$ and $7$ .

$2 (- \frac{1}{1 \sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)})$

Step 20.2.3

Multiply $\sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)$ by $1$ .

$2 (- \frac{1}{\sec^{2} (- 1 \cdot 6 - 1 \cdot 2 \cdot - 3)})$

Step 20.2.4

Simplify each term.

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Step 20.2.4.1

Multiply $- 1$ by $6$ .

$2 (- \frac{1}{\sec^{2} (- 6 - 1 \cdot 2 \cdot - 3)})$

Step 20.2.4.2

Multiply $- 1 \cdot 2 \cdot - 3$ .

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Step 20.2.4.2.1

Multiply $- 1$ by $2$ .

$2 (- \frac{1}{\sec^{2} (- 6 - 2 \cdot - 3)})$

Step 20.2.4.2.2

Multiply $- 2$ by $- 3$ .

$2 (- \frac{1}{\sec^{2} (- 6 + 6)})$

Step 20.2.5

Add $- 6$ and $6$ .

$2 (- \frac{1}{\sec^{2} (0)})$

Step 20.2.6

The exact value of $\sec (0)$ is $1$ .

$2 (- \frac{1}{1^{2}})$

Step 20.2.7

One to any power is one.

$2 (- \frac{1}{1})$

Step 20.3

Cancel the common factor of $1$ .

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Step 20.3.1

Cancel the common factor.

$2 (- \frac{1}{1})$

Step 20.3.2

Rewrite the expression.

$2 (- 1 \cdot 1)$

Step 20.4

Multiply $2 (- 1 \cdot 1)$ .

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Step 20.4.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1$

Step 20.4.2

Multiply $2$ by $- 1$ .

$- 2$