Calculus Examples

$lim_{x \to 0} \frac{\sin (3 x) - \tan (2 x)}{5 x}$

Step 1

Move the term $\frac{1}{5}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{5} lim_{x \to 0} \frac{\sin (3 x) - \tan (2 x)}{x}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

$\frac{1}{5} \cdot \frac{lim_{x \to 0} \sin (3 x) - \tan (2 x)}{lim_{x \to 0} x}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

$\frac{1}{5} \cdot \frac{lim_{x \to 0} \sin (3 x) - lim_{x \to 0} \tan (2 x)}{lim_{x \to 0} x}$

Step 2.1.2.2

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

$\frac{1}{5} \cdot \frac{\sin (lim_{x \to 0} 3 x) - lim_{x \to 0} \tan (2 x)}{lim_{x \to 0} x}$

Step 2.1.2.3

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

$\frac{1}{5} \cdot \frac{\sin (3 lim_{x \to 0} x) - lim_{x \to 0} \tan (2 x)}{lim_{x \to 0} x}$

Step 2.1.2.4

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

$\frac{1}{5} \cdot \frac{\sin (3 lim_{x \to 0} x) - \tan (lim_{x \to 0} 2 x)}{lim_{x \to 0} x}$

Step 2.1.2.5

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

$\frac{1}{5} \cdot \frac{\sin (3 lim_{x \to 0} x) - \tan (2 lim_{x \to 0} x)}{lim_{x \to 0} x}$

Step 2.1.2.6

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

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Step 2.1.2.6.1

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

$\frac{1}{5} \cdot \frac{\sin (3 \cdot 0) - \tan (2 lim_{x \to 0} x)}{lim_{x \to 0} x}$

Step 2.1.2.6.2

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

$\frac{1}{5} \cdot \frac{\sin (3 \cdot 0) - \tan (2 \cdot 0)}{lim_{x \to 0} x}$

Step 2.1.2.7

Simplify the answer.

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Step 2.1.2.7.1

Simplify each term.

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Step 2.1.2.7.1.1

Multiply $3$ by $0$ .

$\frac{1}{5} \cdot \frac{\sin (0) - \tan (2 \cdot 0)}{lim_{x \to 0} x}$

Step 2.1.2.7.1.2

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

$\frac{1}{5} \cdot \frac{0 - \tan (2 \cdot 0)}{lim_{x \to 0} x}$

Step 2.1.2.7.1.3

Multiply $2$ by $0$ .

$\frac{1}{5} \cdot \frac{0 - \tan (0)}{lim_{x \to 0} x}$

Step 2.1.2.7.1.4

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

$\frac{1}{5} \cdot \frac{0 - 0}{lim_{x \to 0} x}$

Step 2.1.2.7.1.5

Multiply $- 1$ by $0$ .

$\frac{1}{5} \cdot \frac{0 + 0}{lim_{x \to 0} x}$

Step 2.1.2.7.2

Add $0$ and $0$ .

$\frac{1}{5} \cdot \frac{0}{lim_{x \to 0} x}$

Step 2.1.3

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

$\frac{1}{5} \cdot \frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{1}{5} \cdot \frac{0}{0}$

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$\frac{1}{5} lim_{x \to 0} \frac{\frac{d}{d x} [\sin (3 x) - \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.2

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

$\frac{1}{5} lim_{x \to 0} \frac{\frac{d}{d x} [\sin (3 x)] + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3

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

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Step 2.3.3.1

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

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Step 2.3.3.1.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$\frac{1}{5} lim_{x \to 0} \frac{\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [3 x] + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.1.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$\frac{1}{5} lim_{x \to 0} \frac{\cos (u_{1}) \frac{d}{d x} [3 x] + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.1.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$\frac{1}{5} lim_{x \to 0} \frac{\cos (3 x) \frac{d}{d x} [3 x] + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.2

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

$\frac{1}{5} lim_{x \to 0} \frac{\cos (3 x) \cdot 3 \frac{d}{d x} [x] + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.3

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

$\frac{1}{5} lim_{x \to 0} \frac{\cos (3 x) \cdot 3 \cdot 1 + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.4

Multiply $3$ by $1$ .

$\frac{1}{5} lim_{x \to 0} \frac{\cos (3 x) \cdot 3 + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.3.5

Move $3$ to the left of $\cos (3 x)$ .

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) + \frac{d}{d x} [- \tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.4

Evaluate $\frac{d}{d x} [- \tan (2 x)]$ .

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Step 2.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \tan (2 x)$ with respect to $x$ is $- \frac{d}{d x} [\tan (2 x)]$ .

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \frac{d}{d x} [\tan (2 x)]}{\frac{d}{d x} [x]}$

Step 2.3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x$ .

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Step 2.3.4.2.1

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [2 x]}{\frac{d}{d x} [x]}$

Step 2.3.4.2.2

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \sec^{2} (u_{2}) \frac{d}{d x} [2 x]}{\frac{d}{d x} [x]}$

Step 2.3.4.2.3

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \sec^{2} (2 x) \frac{d}{d x} [2 x]}{\frac{d}{d x} [x]}$

Step 2.3.4.3

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \sec^{2} (2 x) \cdot 2 \frac{d}{d x} [x]}{\frac{d}{d x} [x]}$

Step 2.3.4.4

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \sec^{2} (2 x) \cdot 2 \cdot 1}{\frac{d}{d x} [x]}$

Step 2.3.4.5

Multiply $2$ by $1$ .

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - \sec^{2} (2 x) \cdot 2}{\frac{d}{d x} [x]}$

Step 2.3.4.6

Move $2$ to the left of $\sec^{2} (2 x)$ .

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - 1 \cdot 2 \cdot \sec^{2} (2 x)}{\frac{d}{d x} [x]}$

Step 2.3.4.7

Multiply $2$ by $- 1$ .

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - 2 \sec^{2} (2 x)}{\frac{d}{d x} [x]}$

Step 2.3.5

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

$\frac{1}{5} lim_{x \to 0} \frac{3 \cos (3 x) - 2 \sec^{2} (2 x)}{1}$

Step 2.4

Divide $3 \cos (3 x) - 2 \sec^{2} (2 x)$ by $1$ .

$\frac{1}{5} lim_{x \to 0} 3 \cos (3 x) - 2 \sec^{2} (2 x)$

Step 3

Evaluate the limit.

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Step 3.1

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

$\frac{1}{5} (lim_{x \to 0} 3 \cos (3 x) - lim_{x \to 0} 2 \sec^{2} (2 x))$

Step 3.2

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

$\frac{1}{5} (3 lim_{x \to 0} \cos (3 x) - lim_{x \to 0} 2 \sec^{2} (2 x))$

Step 3.3

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

$\frac{1}{5} (3 \cos (lim_{x \to 0} 3 x) - lim_{x \to 0} 2 \sec^{2} (2 x))$

Step 3.4

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

$\frac{1}{5} (3 \cos (3 lim_{x \to 0} x) - lim_{x \to 0} 2 \sec^{2} (2 x))$

Step 3.5

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

$\frac{1}{5} (3 \cos (3 lim_{x \to 0} x) - 2 lim_{x \to 0} \sec^{2} (2 x))$

Step 3.6

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

$\frac{1}{5} (3 \cos (3 lim_{x \to 0} x) - 2 {(lim_{x \to 0} \sec (2 x))}^{2})$

Step 3.7

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

$\frac{1}{5} (3 \cos (3 lim_{x \to 0} x) - 2 \sec^{2} (lim_{x \to 0} 2 x))$

Step 3.8

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

$\frac{1}{5} (3 \cos (3 lim_{x \to 0} x) - 2 \sec^{2} (2 lim_{x \to 0} x))$

Step 4

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

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Step 4.1

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

$\frac{1}{5} (3 \cos (3 \cdot 0) - 2 \sec^{2} (2 lim_{x \to 0} x))$

Step 4.2

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

$\frac{1}{5} (3 \cos (3 \cdot 0) - 2 \sec^{2} (2 \cdot 0))$

Step 5

Simplify the answer.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $3$ by $0$ .

$\frac{1}{5} (3 \cos (0) - 2 \sec^{2} (2 \cdot 0))$

Step 5.1.2

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

$\frac{1}{5} (3 \cdot 1 - 2 \sec^{2} (2 \cdot 0))$

Step 5.1.3

Multiply $3$ by $1$ .

$\frac{1}{5} (3 - 2 \sec^{2} (2 \cdot 0))$

Step 5.1.4

Multiply $2$ by $0$ .

$\frac{1}{5} (3 - 2 \sec^{2} (0))$

Step 5.1.5

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

$\frac{1}{5} (3 - 2 \cdot 1^{2})$

Step 5.1.6

One to any power is one.

$\frac{1}{5} (3 - 2 \cdot 1)$

Step 5.1.7

Multiply $- 2$ by $1$ .

$\frac{1}{5} (3 - 2)$

Step 5.2

Subtract $2$ from $3$ .

$\frac{1}{5} \cdot 1$

Step 5.3

Multiply $\frac{1}{5}$ by $1$ .

$\frac{1}{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{5}$

Decimal Form:

$0.2$