Calculus Examples

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

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.

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

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 $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.2.1.2

Move the limit inside the logarithm.

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

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.2.1.4

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.2.1.5

Evaluate the limit of $4$ which is constant as $x$ approaches $- 1$ .

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

Step 1.2.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Multiply $3$ by $- 1$ .

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

Step 1.2.3.2

Add $- 3$ and $4$ .

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

Step 1.2.3.3

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

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

Step 1.2.3.4

Multiply $3$ by $0$ .

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

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 $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.3.1.2

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

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

Step 1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.3.1.4

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

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

Step 1.3.1.5

Evaluate the limit of $2$ which is constant as $x$ approaches $- 1$ .

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

Step 1.3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Multiply $2$ by $- 1$ .

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

Step 1.3.3.2

Add $- 2$ and $2$ .

$\frac{0}{3 \tan (0)}$

Step 1.3.3.3

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

$\frac{0}{3 \cdot 0}$

Step 1.3.3.4

Multiply $3$ 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

Since $3$ is constant with respect to $x$ , the derivative of $3 \ln (3 x + 4)$ with respect to $x$ is $3 \frac{d}{d x} [\ln (3 x + 4)]$ .

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

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 $g (x) = 3 x + 4$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $3 x + 4$ .

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

Step 3.3.2

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

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

Step 3.3.3

Replace all occurrences of $u$ with $3 x + 4$ .

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

Step 3.4

Combine $\frac{1}{3 x + 4}$ and $3$ .

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

Step 3.5

By the Sum Rule, the derivative of $3 x + 4$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [4]$ .

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

Step 3.6

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 3.7

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

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

Step 3.8

Multiply $3$ by $1$ .

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

Step 3.9

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{\frac{3}{3 x + 4} (3 + 0)}{\frac{d}{d x} [3 \tan (2 x + 2)]}$

Step 3.10

Add $3$ and $0$ .

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

Step 3.11

Combine $\frac{3}{3 x + 4}$ and $3$ .

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

Step 3.12

Multiply $3$ by $3$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{\frac{d}{d x} [3 \tan (2 x + 2)]}$

Step 3.13

Since $3$ is constant with respect to $x$ , the derivative of $3 \tan (2 x + 2)$ with respect to $x$ is $3 \frac{d}{d x} [\tan (2 x + 2)]$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \frac{d}{d x} [\tan (2 x + 2)]}$

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 $g (x) = 2 x + 2$ .

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Step 3.14.1

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 (\frac{d}{d u} [\tan (u)] \frac{d}{d x} [2 x + 2])}$

Step 3.14.2

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 (\sec^{2} (u) \frac{d}{d x} [2 x + 2])}$

Step 3.14.3

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 (\sec^{2} (2 x + 2) \frac{d}{d x} [2 x + 2])}$

Step 3.15

Remove parentheses.

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) \frac{d}{d x} [2 x + 2]}$

Step 3.16

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) (\frac{d}{d x} [2 x] + \frac{d}{d x} [2])}$

Step 3.17

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) (2 \frac{d}{d x} [x] + \frac{d}{d x} [2])}$

Step 3.18

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

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) (2 \cdot 1 + \frac{d}{d x} [2])}$

Step 3.19

Multiply $2$ by $1$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) (2 + \frac{d}{d x} [2])}$

Step 3.20

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) (2 + 0)}$

Step 3.21

Add $2$ and $0$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{3 \sec^{2} (2 x + 2) \cdot 2}$

Step 3.22

Multiply $2$ by $3$ .

$lim_{x \to - 1} \frac{\frac{9}{3 x + 4}}{6 \sec^{2} (2 x + 2)}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 1} \frac{9}{3 x + 4} \cdot \frac{1}{6 \sec^{2} (2 x + 2)}$

Step 5

Simplify terms.

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Step 5.1

Multiply $\frac{9}{3 x + 4}$ by $\frac{1}{6 \sec^{2} (2 x + 2)}$ .

$lim_{x \to - 1} \frac{9}{(3 x + 4) (6 \sec^{2} (2 x + 2))}$

Step 5.2

Cancel the common factor of $9$ and $6$ .

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Step 5.2.1

Factor $3$ out of $9$ .

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

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $3$ out of $(3 x + 4) (6 \sec^{2} (2 x + 2))$ .

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

Step 5.2.2.2

Cancel the common factor.

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

Step 5.2.2.3

Rewrite the expression.

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

Step 6

Move the term $\frac{3}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 7

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 1$ .

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

Step 8

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

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

Step 9

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 10

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 11

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 12

Evaluate the limit of $4$ which is constant as $x$ approaches $- 1$ .

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

Step 13

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

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

Step 14

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

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

Step 15

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 16

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

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

Step 17

Evaluate the limit of $2$ which is constant as $x$ approaches $- 1$ .

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

Step 18

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 18.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 18.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{3}{2} \cdot \frac{1}{(3 \cdot - 1 + 4) \cdot \sec^{2} (2 \cdot - 1 + 2)}$

Step 19

Simplify the answer.

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Step 19.1

Combine.

$\frac{3 \cdot 1}{2 ((3 \cdot - 1 + 4) \cdot \sec^{2} (2 \cdot - 1 + 2))}$

Step 19.2

Multiply $3$ by $1$ .

$\frac{3}{2 (3 \cdot - 1 + 4) \cdot \sec^{2} (2 \cdot - 1 + 2)}$

Step 19.3

Simplify the denominator.

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Step 19.3.1

Combine exponents.

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Step 19.3.1.1

Multiply $3$ by $- 1$ .

$\frac{3}{2 (- 3 + 4) \sec^{2} (2 \cdot - 1 + 2)}$

Step 19.3.1.2

Multiply $2$ by $- 1$ .

$\frac{3}{2 (- 3 + 4) \sec^{2} (- 2 + 2)}$

Step 19.3.2

Add $- 3$ and $4$ .

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

Step 19.3.3

Add $- 2$ and $2$ .

$\frac{3}{2 \cdot 1 \sec^{2} (0)}$

Step 19.3.4

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

$\frac{3}{2 \cdot 1 \cdot 1^{2}}$

Step 19.3.5

One to any power is one.

$\frac{3}{2 \cdot 1 \cdot 1}$

Step 19.3.6

Combine exponents.

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Step 19.3.6.1

Multiply $2$ by $1$ .

$\frac{3}{2 \cdot 1}$

Step 19.3.6.2

Multiply $2$ by $1$ .

$\frac{3}{2}$