Calculus Examples

$lim_{x \to 0} \frac{6 e^{4 x} - 2 e^{3 x} - 4}{\sin (2 x)}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{x \to 0} 6 e^{4 x} - 2 e^{3 x} - 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 0} 6 e^{4 x} - lim_{x \to 0} 2 e^{3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.2

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{6 lim_{x \to 0} e^{4 x} - lim_{x \to 0} 2 e^{3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.3

Move the limit into the exponent.

$\frac{6 e^{lim_{x \to 0} 4 x} - lim_{x \to 0} 2 e^{3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.4

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

$\frac{6 e^{4 lim_{x \to 0} x} - lim_{x \to 0} 2 e^{3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.5

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

$\frac{6 e^{4 lim_{x \to 0} x} - 2 lim_{x \to 0} e^{3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.6

Move the limit into the exponent.

$\frac{6 e^{4 lim_{x \to 0} x} - 2 e^{lim_{x \to 0} 3 x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.7

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

$\frac{6 e^{4 lim_{x \to 0} x} - 2 e^{3 lim_{x \to 0} x} - lim_{x \to 0} 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.8

Evaluate the limit of $4$ which is constant as $x$ approaches $0$ .

$\frac{6 e^{4 lim_{x \to 0} x} - 2 e^{3 lim_{x \to 0} x} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.9

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

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Step 1.1.2.9.1

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

$\frac{6 e^{4 \cdot 0} - 2 e^{3 lim_{x \to 0} x} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.9.2

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

$\frac{6 e^{4 \cdot 0} - 2 e^{3 \cdot 0} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10

Simplify the answer.

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Step 1.1.2.10.1

Simplify each term.

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Step 1.1.2.10.1.1

Multiply $4$ by $0$ .

$\frac{6 e^{0} - 2 e^{3 \cdot 0} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.2

Anything raised to $0$ is $1$ .

$\frac{6 \cdot 1 - 2 e^{3 \cdot 0} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.3

Multiply $6$ by $1$ .

$\frac{6 - 2 e^{3 \cdot 0} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.4

Multiply $3$ by $0$ .

$\frac{6 - 2 e^{0} - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.5

Anything raised to $0$ is $1$ .

$\frac{6 - 2 \cdot 1 - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.6

Multiply $- 2$ by $1$ .

$\frac{6 - 2 - 1 \cdot 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.1.7

Multiply $- 1$ by $4$ .

$\frac{6 - 2 - 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.2

Subtract $2$ from $6$ .

$\frac{4 - 4}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.2.10.3

Subtract $4$ from $4$ .

$\frac{0}{lim_{x \to 0} \sin (2 x)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

$\frac{0}{\sin (lim_{x \to 0} 2 x)}$

Step 1.1.3.1.2

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

$\frac{0}{\sin (2 lim_{x \to 0} x)}$

Step 1.1.3.2

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

$\frac{0}{\sin (2 \cdot 0)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Multiply $2$ by $0$ .

$\frac{0}{\sin (0)}$

Step 1.1.3.3.2

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

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 0} \frac{6 e^{4 x} - 2 e^{3 x} - 4}{\sin (2 x)} = lim_{x \to 0} \frac{\frac{d}{d x} [6 e^{4 x} - 2 e^{3 x} - 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [6 e^{4 x} - 2 e^{3 x} - 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.2

By the Sum Rule, the derivative of $6 e^{4 x} - 2 e^{3 x} - 4$ with respect to $x$ is $\frac{d}{d x} [6 e^{4 x}] + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]$ .

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

Step 1.3.3

Evaluate $\frac{d}{d x} [6 e^{4 x}]$ .

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Step 1.3.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 e^{4 x}$ with respect to $x$ is $6 \frac{d}{d x} [e^{4 x}]$ .

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

Step 1.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 1.3.3.2.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$lim_{x \to 0} \frac{6 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x]) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{6 (e^{u_{1}} \frac{d}{d x} [4 x]) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.3.2.3

Replace all occurrences of $u_{1}$ with $4 x$ .

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

Step 1.3.3.3

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

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

Step 1.3.3.4

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

$lim_{x \to 0} \frac{6 (e^{4 x} (4 \cdot 1)) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.3.5

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{6 (e^{4 x} \cdot 4) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.3.6

Move $4$ to the left of $e^{4 x}$ .

$lim_{x \to 0} \frac{6 (4 \cdot e^{4 x}) + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.3.7

Multiply $4$ by $6$ .

$lim_{x \to 0} \frac{24 e^{4 x} + \frac{d}{d x} [- 2 e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 2 e^{3 x}]$ .

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Step 1.3.4.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{3 x}$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{3 x}]$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 \frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 1.3.4.2.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (e^{u_{2}} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.2.3

Replace all occurrences of $u_{2}$ with $3 x$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (e^{3 x} \frac{d}{d x} [3 x]) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.3

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

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (e^{3 x} (3 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.4

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

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.5

Multiply $3$ by $1$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (e^{3 x} \cdot 3) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.6

Move $3$ to the left of $e^{3 x}$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 2 (3 \cdot e^{3 x}) + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.4.7

Multiply $3$ by $- 2$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x} + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.5

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x} + 0}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.6

Add $24 e^{4 x} - 6 e^{3 x}$ and $0$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\frac{d}{d x} [\sin (2 x)]}$

Step 1.3.7

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 $g (x) = 2 x$ .

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Step 1.3.7.1

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x]}$

Step 1.3.7.2

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\cos (u) \frac{d}{d x} [2 x]}$

Step 1.3.7.3

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\cos (2 x) \frac{d}{d x} [2 x]}$

Step 1.3.8

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\cos (2 x) (2 \frac{d}{d x} [x])}$

Step 1.3.9

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

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\cos (2 x) (2 \cdot 1)}$

Step 1.3.10

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{\cos (2 x) \cdot 2}$

Step 1.3.11

Move $2$ to the left of $\cos (2 x)$ .

$lim_{x \to 0} \frac{24 e^{4 x} - 6 e^{3 x}}{2 \cos (2 x)}$

Step 1.4

Cancel the common factor of $24 e^{4 x} - 6 e^{3 x}$ and $2$ .

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Step 1.4.1

Factor $2$ out of $24 e^{4 x}$ .

$lim_{x \to 0} \frac{2 (12 e^{4 x}) - 6 e^{3 x}}{2 \cos (2 x)}$

Step 1.4.2

Factor $2$ out of $- 6 e^{3 x}$ .

$lim_{x \to 0} \frac{2 (12 e^{4 x}) + 2 (- 3 e^{3 x})}{2 \cos (2 x)}$

Step 1.4.3

Factor $2$ out of $2 (12 e^{4 x}) + 2 (- 3 e^{3 x})$ .

$lim_{x \to 0} \frac{2 (12 e^{4 x} - 3 e^{3 x})}{2 \cos (2 x)}$

Step 1.4.4

Cancel the common factors.

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Step 1.4.4.1

Factor $2$ out of $2 \cos (2 x)$ .

$lim_{x \to 0} \frac{2 (12 e^{4 x} - 3 e^{3 x})}{2 (\cos (2 x))}$

Step 1.4.4.2

Cancel the common factor.

$lim_{x \to 0} \frac{2 (12 e^{4 x} - 3 e^{3 x})}{2 \cos (2 x)}$

Step 1.4.4.3

Rewrite the expression.

$lim_{x \to 0} \frac{12 e^{4 x} - 3 e^{3 x}}{\cos (2 x)}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{lim_{x \to 0} 12 e^{4 x} - 3 e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.2

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

$\frac{lim_{x \to 0} 12 e^{4 x} - lim_{x \to 0} 3 e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.3

Move the term $12$ outside of the limit because it is constant with respect to $x$ .

$\frac{12 lim_{x \to 0} e^{4 x} - lim_{x \to 0} 3 e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.4

Move the limit into the exponent.

$\frac{12 e^{lim_{x \to 0} 4 x} - lim_{x \to 0} 3 e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.5

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

$\frac{12 e^{4 lim_{x \to 0} x} - lim_{x \to 0} 3 e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.6

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

$\frac{12 e^{4 lim_{x \to 0} x} - 3 lim_{x \to 0} e^{3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.7

Move the limit into the exponent.

$\frac{12 e^{4 lim_{x \to 0} x} - 3 e^{lim_{x \to 0} 3 x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.8

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

$\frac{12 e^{4 lim_{x \to 0} x} - 3 e^{3 lim_{x \to 0} x}}{lim_{x \to 0} \cos (2 x)}$

Step 2.9

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

$\frac{12 e^{4 lim_{x \to 0} x} - 3 e^{3 lim_{x \to 0} x}}{\cos (lim_{x \to 0} 2 x)}$

Step 2.10

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

$\frac{12 e^{4 lim_{x \to 0} x} - 3 e^{3 lim_{x \to 0} x}}{\cos (2 lim_{x \to 0} x)}$

Step 3

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

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Step 3.1

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

$\frac{12 e^{4 \cdot 0} - 3 e^{3 lim_{x \to 0} x}}{\cos (2 lim_{x \to 0} x)}$

Step 3.2

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

$\frac{12 e^{4 \cdot 0} - 3 e^{3 \cdot 0}}{\cos (2 lim_{x \to 0} x)}$

Step 3.3

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

$\frac{12 e^{4 \cdot 0} - 3 e^{3 \cdot 0}}{\cos (2 \cdot 0)}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $4$ by $0$ .

$\frac{12 e^{0} - 3 e^{3 \cdot 0}}{\cos (2 \cdot 0)}$

Step 4.1.2

Anything raised to $0$ is $1$ .

$\frac{12 \cdot 1 - 3 e^{3 \cdot 0}}{\cos (2 \cdot 0)}$

Step 4.1.3

Multiply $12$ by $1$ .

$\frac{12 - 3 e^{3 \cdot 0}}{\cos (2 \cdot 0)}$

Step 4.1.4

Multiply $3$ by $0$ .

$\frac{12 - 3 e^{0}}{\cos (2 \cdot 0)}$

Step 4.1.5

Anything raised to $0$ is $1$ .

$\frac{12 - 3 \cdot 1}{\cos (2 \cdot 0)}$

Step 4.1.6

Multiply $- 3$ by $1$ .

$\frac{12 - 3}{\cos (2 \cdot 0)}$

Step 4.1.7

Subtract $3$ from $12$ .

$\frac{9}{\cos (2 \cdot 0)}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $0$ .

$\frac{9}{\cos (0)}$

Step 4.2.2

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

$\frac{9}{1}$

Step 4.3

Divide $9$ by $1$ .

$9$