Calculus Examples

$lim_{x \to 0} \frac{e^{- x} - 1}{3 \sin (- x) + 3 x}$

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.

$\frac{lim_{x \to 0} e^{- x} - 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{lim_{x \to 0} e^{- x} - lim_{x \to 0} 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2.2

Move the limit into the exponent.

$\frac{e^{lim_{x \to 0} - x} - lim_{x \to 0} 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2.3

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

$\frac{e^{- lim_{x \to 0} x} - lim_{x \to 0} 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2.4

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\frac{e^{- lim_{x \to 0} x} - 1 \cdot 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2.5

Simplify terms.

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Step 1.2.5.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{e^{0} - 1 \cdot 1}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

Step 1.2.5.2

Simplify the answer.

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Step 1.2.5.2.1

Simplify each term.

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Step 1.2.5.2.1.1

Anything raised to $0$ is $1$ .

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

Step 1.2.5.2.1.2

Multiply $- 1$ by $1$ .

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

Step 1.2.5.2.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} 3 \sin (- x) + 3 x}$

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 $0$ .

$\frac{0}{lim_{x \to 0} 3 \sin (- x) + lim_{x \to 0} 3 x}$

Step 1.3.2

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

$\frac{0}{3 lim_{x \to 0} \sin (- x) + lim_{x \to 0} 3 x}$

Step 1.3.3

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

$\frac{0}{3 \sin (lim_{x \to 0} - x) + lim_{x \to 0} 3 x}$

Step 1.3.4

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

$\frac{0}{3 \sin (- lim_{x \to 0} x) + lim_{x \to 0} 3 x}$

Step 1.3.5

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

$\frac{0}{3 \sin (- lim_{x \to 0} x) + 3 lim_{x \to 0} x}$

Step 1.3.6

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.3.6.1

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{3 \sin (0) + 3 lim_{x \to 0} x}$

Step 1.3.6.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{3 \sin (0) + 3 \cdot 0}$

Step 1.3.7

Simplify the answer.

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Step 1.3.7.1

Simplify each term.

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Step 1.3.7.1.1

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

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

Step 1.3.7.1.2

Multiply $3$ by $0$ .

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

Step 1.3.7.1.3

Multiply $3$ by $0$ .

$\frac{0}{0 + 0}$

Step 1.3.7.2

Add $0$ and $0$ .

$\frac{0}{0}$

Step 1.3.7.3

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

Undefined

$\frac{0}{0}$

Step 1.3.8

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.

$lim_{x \to 0} \frac{e^{- x} - 1}{3 \sin (- x) + 3 x} = lim_{x \to 0} \frac{\frac{d}{d x} [e^{- x} - 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{- x} - 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.2

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

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{- x}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3

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

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Step 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 $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 3.3.1.1

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

$lim_{x \to 0} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.1.2

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

$lim_{x \to 0} \frac{e^{u} \frac{d}{d x} [- x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.1.3

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

$lim_{x \to 0} \frac{e^{- x} \frac{d}{d x} [- x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{e^{- x} (- \frac{d}{d x} [x]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.3

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

$lim_{x \to 0} \frac{e^{- x} (- 1 \cdot 1) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.4

Multiply $- 1$ by $1$ .

$lim_{x \to 0} \frac{e^{- x} \cdot - 1 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.5

Move $- 1$ to the left of $e^{- x}$ .

$lim_{x \to 0} \frac{- 1 e^{- x} + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.3.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$lim_{x \to 0} \frac{- e^{- x} + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.4

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

$lim_{x \to 0} \frac{- e^{- x} + 0}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.5

Add $- e^{- x}$ and $0$ .

$lim_{x \to 0} \frac{- e^{- x}}{\frac{d}{d x} [3 \sin (- x) + 3 x]}$

Step 3.6

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

$lim_{x \to 0} \frac{- e^{- x}}{\frac{d}{d x} [3 \sin (- x)] + \frac{d}{d x} [3 x]}$

Step 3.7

Evaluate $\frac{d}{d x} [3 \sin (- x)]$ .

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Step 3.7.1

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

$lim_{x \to 0} \frac{- e^{- x}}{3 \frac{d}{d x} [\sin (- x)] + \frac{d}{d x} [3 x]}$

Step 3.7.2

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

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Step 3.7.2.1

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

$lim_{x \to 0} \frac{- e^{- x}}{3 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [- x]) + \frac{d}{d x} [3 x]}$

Step 3.7.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 (\cos (u) \frac{d}{d x} [- x]) + \frac{d}{d x} [3 x]}$

Step 3.7.2.3

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

$lim_{x \to 0} \frac{- e^{- x}}{3 (\cos (- x) \frac{d}{d x} [- x]) + \frac{d}{d x} [3 x]}$

Step 3.7.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 (\cos (- x) (- \frac{d}{d x} [x])) + \frac{d}{d x} [3 x]}$

Step 3.7.4

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

$lim_{x \to 0} \frac{- e^{- x}}{3 (\cos (- x) (- 1 \cdot 1)) + \frac{d}{d x} [3 x]}$

Step 3.7.5

Multiply $- 1$ by $1$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 (\cos (- x) \cdot - 1) + \frac{d}{d x} [3 x]}$

Step 3.7.6

Move $- 1$ to the left of $\cos (- x)$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 (- 1 \cos (- x)) + \frac{d}{d x} [3 x]}$

Step 3.7.7

Rewrite $- 1 \cos (- x)$ as $- \cos (- x)$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 (- \cos (- x)) + \frac{d}{d x} [3 x]}$

Step 3.7.8

Multiply $- 1$ by $3$ .

$lim_{x \to 0} \frac{- e^{- x}}{- 3 \cos (- x) + \frac{d}{d x} [3 x]}$

Step 3.8

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 3.8.1

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

$lim_{x \to 0} \frac{- e^{- x}}{- 3 \cos (- x) + 3 \frac{d}{d x} [x]}$

Step 3.8.2

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

$lim_{x \to 0} \frac{- e^{- x}}{- 3 \cos (- x) + 3 \cdot 1}$

Step 3.8.3

Multiply $3$ by $1$ .

$lim_{x \to 0} \frac{- e^{- x}}{- 3 \cos (- x) + 3}$

Step 3.9

Simplify.

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Step 3.9.1

Reorder terms.

$lim_{x \to 0} \frac{- e^{- x}}{3 - 3 \cos (- x)}$

Step 3.9.2

Since $\cos (- x)$ is an even function, rewrite $\cos (- x)$ as $\cos (x)$ .

$lim_{x \to 0} \frac{- e^{- x}}{3 - 3 \cos (x)}$

Step 4

Since the numerator is negative and the denominator $3 - 3 \cos (x)$ approaches zero and is greater than zero for $x$ near $0$ on both sides, the function decreases without bound.

$- \infty$