Calculus Examples

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

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.

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

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 $0$ .

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

Step 1.2.2

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

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

Step 1.2.3

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

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

Step 1.2.4

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

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Step 1.2.4.1

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

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

Step 1.2.4.2

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

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

Step 1.2.5

Simplify the answer.

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Step 1.2.5.1

Simplify each term.

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Step 1.2.5.1.1

Raising $0$ to any positive power yields $0$ .

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

Step 1.2.5.1.2

Multiply $3$ by $0$ .

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

Step 1.2.5.2

Add $0$ and $0$ .

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

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 $0$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

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 $x$ .

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

Step 1.3.6.2

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

$\frac{0}{4 \sin (2 \cdot 0) - 3 \cdot 0}$

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

Multiply $2$ by $0$ .

$\frac{0}{4 \sin (0) - 3 \cdot 0}$

Step 1.3.7.1.2

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

$\frac{0}{4 \cdot 0 - 3 \cdot 0}$

Step 1.3.7.1.3

Multiply $4$ by $0$ .

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

Step 1.3.7.1.4

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 3.3.3

Multiply $2$ by $3$ .

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Evaluate $\frac{d}{d x} [4 \sin (2 x)]$ .

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Step 3.6.1

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

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

Step 3.6.2

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 $g (x) = 2 x$ .

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Step 3.6.2.1

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

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

Step 3.6.2.2

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

$lim_{x \to 0} \frac{6 x + 1}{4 (\cos (u) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 3 x]}$

Step 3.6.2.3

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

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

Step 3.6.3

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

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

Step 3.6.4

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

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

Step 3.6.5

Multiply $2$ by $1$ .

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

Step 3.6.6

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

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

Step 3.6.7

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{6 x + 1}{8 \cos (2 x) + \frac{d}{d x} [- 3 x]}$

Step 3.7

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

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Step 3.7.1

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

$lim_{x \to 0} \frac{6 x + 1}{8 \cos (2 x) - 3 \frac{d}{d x} [x]}$

Step 3.7.2

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

$lim_{x \to 0} \frac{6 x + 1}{8 \cos (2 x) - 3 \cdot 1}$

Step 3.7.3

Multiply $- 3$ by $1$ .

$lim_{x \to 0} \frac{6 x + 1}{8 \cos (2 x) - 3}$

Step 3.8

Reorder terms.

$lim_{x \to 0} \frac{6 x + 1}{- 3 + 8 \cos (2 x)}$

Step 4

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

$\frac{lim_{x \to 0} 6 x + 1}{lim_{x \to 0} - 3 + 8 \cos (2 x)}$

Step 5

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

$\frac{lim_{x \to 0} 6 x + lim_{x \to 0} 1}{lim_{x \to 0} - 3 + 8 \cos (2 x)}$

Step 6

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

$\frac{6 lim_{x \to 0} x + lim_{x \to 0} 1}{lim_{x \to 0} - 3 + 8 \cos (2 x)}$

Step 7

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

$\frac{6 lim_{x \to 0} x + 1}{lim_{x \to 0} - 3 + 8 \cos (2 x)}$

Step 8

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

$\frac{6 lim_{x \to 0} x + 1}{- lim_{x \to 0} 3 + lim_{x \to 0} 8 \cos (2 x)}$

Step 9

Evaluate the limit of $3$ which is constant as $x$ approaches $0$ .

$\frac{6 lim_{x \to 0} x + 1}{- 1 \cdot 3 + lim_{x \to 0} 8 \cos (2 x)}$

Step 10

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

$\frac{6 lim_{x \to 0} x + 1}{- 3 + 8 lim_{x \to 0} \cos (2 x)}$

Step 11

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

$\frac{6 lim_{x \to 0} x + 1}{- 3 + 8 \cos (lim_{x \to 0} 2 x)}$

Step 12

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

$\frac{6 lim_{x \to 0} x + 1}{- 3 + 8 \cos (2 lim_{x \to 0} x)}$

Step 13

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

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Step 13.1

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

$\frac{6 \cdot 0 + 1}{- 3 + 8 \cos (2 lim_{x \to 0} x)}$

Step 13.2

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

$\frac{6 \cdot 0 + 1}{- 3 + 8 \cos (2 \cdot 0)}$

Step 14

Simplify the answer.

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Step 14.1

Simplify the numerator.

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Step 14.1.1

Multiply $6$ by $0$ .

$\frac{0 + 1}{- 3 + 8 \cos (2 \cdot 0)}$

Step 14.1.2

Add $0$ and $1$ .

$\frac{1}{- 3 + 8 \cos (2 \cdot 0)}$

Step 14.2

Simplify the denominator.

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Step 14.2.1

Multiply $2$ by $0$ .

$\frac{1}{- 3 + 8 \cos (0)}$

Step 14.2.2

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

$\frac{1}{- 3 + 8 \cdot 1}$

Step 14.2.3

Multiply $8$ by $1$ .

$\frac{1}{- 3 + 8}$

Step 14.2.4

Add $- 3$ and $8$ .

$\frac{1}{5}$