Calculus Examples

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

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.

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

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 $x$ .

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

Step 1.2.1.2

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

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

Step 1.2.1.3

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

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

Step 1.2.1.4

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

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

Step 1.2.1.5

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

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

Step 1.2.2

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

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Multiply $- 1$ by $2$ .

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

Step 1.2.3.1.2

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

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Step 1.2.3.1.2.1

Multiply $- 1$ by $2$ .

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

Step 1.2.3.1.2.2

Multiply $- 2$ by $- 1$ .

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

Step 1.2.3.2

Add $- 2$ and $2$ .

$\frac{4 \tan (0)}{lim_{x \to - 1} e^{x + 1} + x}$

Step 1.2.3.3

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

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

Step 1.2.3.4

Multiply $4$ by $0$ .

$\frac{0}{lim_{x \to - 1} e^{x + 1} + x}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

$\frac{0}{lim_{x \to - 1} e^{x + 1} + lim_{x \to - 1} x}$

Step 1.3.2

Move the limit into the exponent.

$\frac{0}{e^{lim_{x \to - 1} x + 1} + lim_{x \to - 1} x}$

Step 1.3.3

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

$\frac{0}{e^{lim_{x \to - 1} x + lim_{x \to - 1} 1} + lim_{x \to - 1} x}$

Step 1.3.4

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

$\frac{0}{e^{lim_{x \to - 1} x + 1} + lim_{x \to - 1} x}$

Step 1.3.5

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

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Step 1.3.5.1

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

$\frac{0}{e^{- 1 + 1} + lim_{x \to - 1} x}$

Step 1.3.5.2

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

$\frac{0}{e^{- 1 + 1} - 1}$

Step 1.3.6

Simplify the answer.

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Step 1.3.6.1

Simplify each term.

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Step 1.3.6.1.1

Add $- 1$ and $1$ .

$\frac{0}{e^{0} - 1}$

Step 1.3.6.1.2

Anything raised to $0$ is $1$ .

$\frac{0}{1 - 1}$

Step 1.3.6.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.3.6.3

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

Undefined

$\frac{0}{0}$

Step 1.3.7

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

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 $f (x) = \tan (x)$ and $g (x) = - 2 - 2 x$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

Remove parentheses.

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

Step 3.5

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

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

Step 3.6

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

$lim_{x \to - 1} \frac{4 \sec^{2} (- 2 - 2 x) (0 + \frac{d}{d x} [- 2 x])}{\frac{d}{d x} [e^{x + 1} + x]}$

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Multiply $- 2$ by $4$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x) \frac{d}{d x} [x]}{\frac{d}{d x} [e^{x + 1} + x]}$

Step 3.10

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

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x) \cdot 1}{\frac{d}{d x} [e^{x + 1} + x]}$

Step 3.11

Multiply $- 8$ by $1$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{\frac{d}{d x} [e^{x + 1} + x]}$

Step 3.12

By the Sum Rule, the derivative of $e^{x + 1} + x$ with respect to $x$ is $\frac{d}{d x} [e^{x + 1}] + \frac{d}{d x} [x]$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{\frac{d}{d x} [e^{x + 1}] + \frac{d}{d x} [x]}$

Step 3.13

Evaluate $\frac{d}{d x} [e^{x + 1}]$ .

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Step 3.13.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x + 1$ .

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Step 3.13.1.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [x + 1] + \frac{d}{d x} [x]}$

Step 3.13.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{u} \frac{d}{d x} [x + 1] + \frac{d}{d x} [x]}$

Step 3.13.1.3

Replace all occurrences of $u$ with $x + 1$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} \frac{d}{d x} [x + 1] + \frac{d}{d x} [x]}$

Step 3.13.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [x]}$

Step 3.13.3

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

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} (1 + \frac{d}{d x} [1]) + \frac{d}{d x} [x]}$

Step 3.13.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} (1 + 0) + \frac{d}{d x} [x]}$

Step 3.13.5

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} \cdot 1 + \frac{d}{d x} [x]}$

Step 3.13.6

Multiply $e^{x + 1}$ by $1$ .

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} + \frac{d}{d x} [x]}$

Step 3.14

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

$lim_{x \to - 1} \frac{- 8 \sec^{2} (- 2 - 2 x)}{e^{x + 1} + 1}$

Step 4

Move the term $- 8$ outside of the limit because it is constant with respect to $x$ .

$- 8 lim_{x \to - 1} \frac{\sec^{2} (- 2 - 2 x)}{e^{x + 1} + 1}$

Step 5

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

$- 8 \frac{lim_{x \to - 1} \sec^{2} (- 2 - 2 x)}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 6

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

$- 8 \frac{{(lim_{x \to - 1} \sec (- 2 - 2 x))}^{2}}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 7

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

$- 8 \frac{\sec^{2} (lim_{x \to - 1} - 2 - 2 x)}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 8

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

$- 8 \frac{\sec^{2} (- lim_{x \to - 1} 2 - lim_{x \to - 1} 2 x)}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 9

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - lim_{x \to - 1} 2 x)}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 10

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{lim_{x \to - 1} e^{x + 1} + 1}$

Step 11

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{lim_{x \to - 1} e^{x + 1} + lim_{x \to - 1} 1}$

Step 12

Move the limit into the exponent.

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{e^{lim_{x \to - 1} x + 1} + lim_{x \to - 1} 1}$

Step 13

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{e^{lim_{x \to - 1} x + lim_{x \to - 1} 1} + lim_{x \to - 1} 1}$

Step 14

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{e^{lim_{x \to - 1} x + 1} + lim_{x \to - 1} 1}$

Step 15

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 lim_{x \to - 1} x)}{e^{lim_{x \to - 1} x + 1} + 1}$

Step 16

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

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Step 16.1

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 \cdot - 1)}{e^{lim_{x \to - 1} x + 1} + 1}$

Step 16.2

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

$- 8 \frac{\sec^{2} (- 1 \cdot 2 - 1 \cdot 2 \cdot - 1)}{e^{- 1 + 1} + 1}$

Step 17

Simplify the answer.

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Step 17.1

Simplify the numerator.

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Step 17.1.1

Simplify each term.

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Step 17.1.1.1

Multiply $- 1$ by $2$ .

$- 8 \frac{\sec^{2} (- 2 - 1 \cdot 2 \cdot - 1)}{e^{- 1 + 1} + 1}$

Step 17.1.1.2

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

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Step 17.1.1.2.1

Multiply $- 1$ by $2$ .

$- 8 \frac{\sec^{2} (- 2 - 2 \cdot - 1)}{e^{- 1 + 1} + 1}$

Step 17.1.1.2.2

Multiply $- 2$ by $- 1$ .

$- 8 \frac{\sec^{2} (- 2 + 2)}{e^{- 1 + 1} + 1}$

Step 17.1.2

Add $- 2$ and $2$ .

$- 8 \frac{\sec^{2} (0)}{e^{- 1 + 1} + 1}$

Step 17.1.3

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

$- 8 \frac{1^{2}}{e^{- 1 + 1} + 1}$

Step 17.1.4

One to any power is one.

$- 8 \frac{1}{e^{- 1 + 1} + 1}$

Step 17.2

Simplify the denominator.

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Step 17.2.1

Add $- 1$ and $1$ .

$- 8 \frac{1}{e^{0} + 1}$

Step 17.2.2

Anything raised to $0$ is $1$ .

$- 8 \frac{1}{1 + 1}$

Step 17.2.3

Add $1$ and $1$ .

$- 8 (\frac{1}{2})$

Step 17.3

Cancel the common factor of $2$ .

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Step 17.3.1

Factor $2$ out of $- 8$ .

$2 (- 4) \frac{1}{2}$

Step 17.3.2

Cancel the common factor.

$2 \cdot - 4 \frac{1}{2}$

Step 17.3.3

Rewrite the expression.

$- 4$