Calculus Examples

$lim_{h \to 0} \frac{\tan (2 (x + h)) - \tan (2 x)}{h}$

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.

$\frac{lim_{h \to 0} \tan (2 (x + h)) - \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

$\frac{lim_{h \to 0} \tan (2 (x + h)) - lim_{h \to 0} \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.1.2

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

$\frac{\tan (lim_{h \to 0} 2 (x + h)) - lim_{h \to 0} \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.1.3

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

$\frac{\tan (2 lim_{h \to 0} x + h) - lim_{h \to 0} \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.1.4

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

$\frac{\tan (2 (lim_{h \to 0} x + lim_{h \to 0} h)) - lim_{h \to 0} \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.1.5

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

$\frac{\tan (2 (x + lim_{h \to 0} h)) - lim_{h \to 0} \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.1.6

Evaluate the limit of $\tan (2 x)$ which is constant as $h$ approaches $0$ .

$\frac{\tan (2 (x + lim_{h \to 0} h)) - \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.2

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

$\frac{\tan (2 (x + 0)) - \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.3

Combine the opposite terms in $\tan (2 (x + 0)) - \tan (2 x)$ .

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Step 1.1.2.3.1

Add $x$ and $0$ .

$\frac{\tan (2 x) - \tan (2 x)}{lim_{h \to 0} h}$

Step 1.1.2.3.2

Subtract $\tan (2 x)$ from $\tan (2 x)$ .

$\frac{0}{lim_{h \to 0} h}$

Step 1.1.3

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

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{h \to 0} \frac{\frac{d}{d h} [\tan (2 (x + h)) - \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.2

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

$lim_{h \to 0} \frac{\frac{d}{d h} [\tan (2 (x + h))] + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3

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

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Step 1.3.3.1

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

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Step 1.3.3.1.1

To apply the Chain Rule, set $u$ as $2 (x + h)$ .

$lim_{h \to 0} \frac{\frac{d}{d u} [\tan (u)] \frac{d}{d h} [2 (x + h)] + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.1.2

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

$lim_{h \to 0} \frac{\sec^{2} (u) \frac{d}{d h} [2 (x + h)] + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.1.3

Replace all occurrences of $u$ with $2 (x + h)$ .

$lim_{h \to 0} \frac{\sec^{2} (2 (x + h)) \frac{d}{d h} [2 (x + h)] + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.2

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

$lim_{h \to 0} \frac{\sec^{2} (2 (x + h)) (2 \frac{d}{d h} [x + h]) + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.3

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

$lim_{h \to 0} \frac{\sec^{2} (2 (x + h)) (2 (\frac{d}{d h} [x] + \frac{d}{d h} [h])) + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.4

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

$lim_{h \to 0} \frac{\sec^{2} (2 (x + h)) (2 (0 + \frac{d}{d h} [h])) + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.5

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

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

Step 1.3.3.6

Add $0$ and $1$ .

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

Step 1.3.3.7

Multiply $2$ by $1$ .

$lim_{h \to 0} \frac{\sec^{2} (2 (x + h)) \cdot 2 + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.3.8

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

$lim_{h \to 0} \frac{2 \sec^{2} (2 (x + h)) + \frac{d}{d h} [- \tan (2 x)]}{\frac{d}{d h} [h]}$

Step 1.3.4

Since $- \tan (2 x)$ is constant with respect to $h$ , the derivative of $- \tan (2 x)$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{2 \sec^{2} (2 (x + h)) + 0}{\frac{d}{d h} [h]}$

Step 1.3.5

Simplify.

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Step 1.3.5.1

Apply the distributive property.

$lim_{h \to 0} \frac{2 \sec^{2} (2 x + 2 h) + 0}{\frac{d}{d h} [h]}$

Step 1.3.5.2

Add $2 \sec^{2} (2 x + 2 h)$ and $0$ .

$lim_{h \to 0} \frac{2 \sec^{2} (2 x + 2 h)}{\frac{d}{d h} [h]}$

Step 1.3.5.3

Rewrite $\sec (2 x + 2 h)$ in terms of sines and cosines.

$lim_{h \to 0} \frac{2 {(\frac{1}{\cos (2 x + 2 h)})}^{2}}{\frac{d}{d h} [h]}$

Step 1.3.5.4

Apply the product rule to $\frac{1}{\cos (2 x + 2 h)}$ .

$lim_{h \to 0} \frac{2 \frac{1^{2}}{\cos^{2} (2 x + 2 h)}}{\frac{d}{d h} [h]}$

Step 1.3.5.5

One to any power is one.

$lim_{h \to 0} \frac{2 \frac{1}{\cos^{2} (2 x + 2 h)}}{\frac{d}{d h} [h]}$

Step 1.3.5.6

Combine $2$ and $\frac{1}{\cos^{2} (2 x + 2 h)}$ .

$lim_{h \to 0} \frac{\frac{2}{\cos^{2} (2 x + 2 h)}}{\frac{d}{d h} [h]}$

Step 1.3.6

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

$lim_{h \to 0} \frac{\frac{2}{\cos^{2} (2 x + 2 h)}}{1}$

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{h \to 0} \frac{2}{\cos^{2} (2 x + 2 h)} \cdot 1$

Step 1.5

Multiply $\frac{2}{\cos^{2} (2 x + 2 h)}$ by $1$ .

$lim_{h \to 0} \frac{2}{\cos^{2} (2 x + 2 h)}$

Step 2

Evaluate the limit.

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Step 2.1

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

$2 lim_{h \to 0} \frac{1}{\cos^{2} (2 x + 2 h)}$

Step 2.2

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

$2 \frac{lim_{h \to 0} 1}{lim_{h \to 0} \cos^{2} (2 x + 2 h)}$

Step 2.3

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

$2 \frac{1}{lim_{h \to 0} \cos^{2} (2 x + 2 h)}$

Step 2.4

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

$2 \frac{1}{{(lim_{h \to 0} \cos (2 x + 2 h))}^{2}}$

Step 2.5

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

$2 \frac{1}{\cos^{2} (lim_{h \to 0} 2 x + 2 h)}$

Step 2.6

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

$2 \frac{1}{\cos^{2} (lim_{h \to 0} 2 x + lim_{h \to 0} 2 h)}$

Step 2.7

Evaluate the limit of $2 x$ which is constant as $h$ approaches $0$ .

$2 \frac{1}{\cos^{2} (2 x + lim_{h \to 0} 2 h)}$

Step 2.8

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

$2 \frac{1}{\cos^{2} (2 x + 2 lim_{h \to 0} h)}$

Step 3

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

$2 \frac{1}{\cos^{2} (2 x + 2 \cdot 0)}$

Step 4

Simplify the answer.

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Step 4.1

Rewrite $1$ as $1^{2}$ .

$2 \frac{1^{2}}{\cos^{2} (2 x + 2 \cdot 0)}$

Step 4.2

Rewrite $\frac{1^{2}}{\cos^{2} (2 x + 2 \cdot 0)}$ as ${(\frac{1}{\cos (2 x + 2 \cdot 0)})}^{2}$ .

$2 {(\frac{1}{\cos (2 x + 2 \cdot 0)})}^{2}$

Step 4.3

Convert from $\frac{1}{\cos (2 x + 2 \cdot 0)}$ to $\sec (2 x + 2 \cdot 0)$ .

$2 \sec^{2} (2 x + 2 \cdot 0)$

Step 4.4

Multiply $2$ by $0$ .

$2 \sec^{2} (2 x + 0)$

Step 4.5

Add $2 x$ and $0$ .

$2 \sec^{2} (2 x)$