Calculus Examples

$lim_{x \to 5} \frac{{(x - 5)}^{3}}{\sin (x - 5) - (x - 5)}$

Step 1

Apply L'Hospital's rule.

Tap for more steps...

Step 1.1

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

Tap for more steps...

Step 1.1.1

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

$\frac{lim_{x \to 5} {(x - 5)}^{3}}{lim_{x \to 5} \sin (x - 5) - (x - 5)}$

Step 1.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.1.2.1

Evaluate the limit.

Tap for more steps...

Step 1.1.2.1.1

Move the exponent $3$ from ${(x - 5)}^{3}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 5} x - 5)}^{3}}{lim_{x \to 5} \sin (x - 5) - (x - 5)}$

Step 1.1.2.1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{{(lim_{x \to 5} x - lim_{x \to 5} 5)}^{3}}{lim_{x \to 5} \sin (x - 5) - (x - 5)}$

Step 1.1.2.1.3

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

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

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

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

Step 1.1.2.3

Simplify the answer.

Tap for more steps...

Step 1.1.2.3.1

Multiply $- 1$ by $5$ .

$\frac{{(5 - 5)}^{3}}{lim_{x \to 5} \sin (x - 5) - (x - 5)}$

Step 1.1.2.3.2

Subtract $5$ from $5$ .

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

Step 1.1.2.3.3

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

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

Step 1.1.3

Evaluate the limits by plugging in $5$ for all occurrences of $x$ .

Tap for more steps...

Step 1.1.3.1

Evaluate the limit of $\sin (x - 5) - (x - 5)$ by plugging in $5$ for $x$ .

$\frac{0}{\sin (5 - 5) - (5 - 5)}$

Step 1.1.3.2

Simplify each term.

Tap for more steps...

Step 1.1.3.2.1

Subtract $5$ from $5$ .

$\frac{0}{\sin (0) - (5 - 5)}$

Step 1.1.3.2.2

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

$\frac{0}{0 - (5 - 5)}$

Step 1.1.3.2.3

Subtract $5$ from $5$ .

$\frac{0}{0 - 0}$

Step 1.1.3.2.4

Multiply $- 1$ by $0$ .

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

Step 1.1.3.3

Add $0$ and $0$ .

$\frac{0}{0}$

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.

$lim_{x \to 5} \frac{{(x - 5)}^{3}}{\sin (x - 5) - (x - 5)} = lim_{x \to 5} \frac{\frac{d}{d x} [{(x - 5)}^{3}]}{\frac{d}{d x} [\sin (x - 5) - (x - 5)]}$

Step 1.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 5} \frac{\frac{d}{d x} [{(x - 5)}^{3}]}{\frac{d}{d x} [\sin (x - 5) - (x - 5)]}$

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x - 5$ .

Tap for more steps...

Step 1.3.2.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$lim_{x \to 5} \frac{\frac{d}{d u} [u^{3}] \frac{d}{d x} [x - 5]}{\frac{d}{d x} [\sin (x - 5) - (x - 5)]}$

Step 1.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$lim_{x \to 5} \frac{3 u^{2} \frac{d}{d x} [x - 5]}{\frac{d}{d x} [\sin (x - 5) - (x - 5)]}$

Step 1.3.2.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 1.3.3

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

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

Step 1.3.4

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

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

Step 1.3.5

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

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

Step 1.3.6

Add $1$ and $0$ .

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

Step 1.3.7

Multiply $3$ by $1$ .

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

Step 1.3.8

By the Sum Rule, the derivative of $\sin (x - 5) - (x - 5)$ with respect to $x$ is $\frac{d}{d x} [\sin (x - 5)] + \frac{d}{d x} [- (x - 5)]$ .

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

Step 1.3.9

Evaluate $\frac{d}{d x} [\sin (x - 5)]$ .

Tap for more steps...

Step 1.3.9.1

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 $g (x) = x - 5$ .

Tap for more steps...

Step 1.3.9.1.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$lim_{x \to 5} \frac{3 {(x - 5)}^{2}}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [x - 5] + \frac{d}{d x} [- (x - 5)]}$

Step 1.3.9.1.2

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

$lim_{x \to 5} \frac{3 {(x - 5)}^{2}}{\cos (u) \frac{d}{d x} [x - 5] + \frac{d}{d x} [- (x - 5)]}$

Step 1.3.9.1.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 1.3.9.2

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

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

Step 1.3.9.3

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

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

Step 1.3.9.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

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

Step 1.3.9.5

Add $1$ and $0$ .

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

Step 1.3.9.6

Multiply $\cos (x - 5)$ by $1$ .

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

Step 1.3.10

Evaluate $\frac{d}{d x} [- (x - 5)]$ .

Tap for more steps...

Step 1.3.10.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- (x - 5)$ with respect to $x$ is $- \frac{d}{d x} [x - 5]$ .

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

Step 1.3.10.2

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

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

Step 1.3.10.3

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

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

Step 1.3.10.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

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

Step 1.3.10.5

Add $1$ and $0$ .

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

Step 1.3.10.6

Multiply $- 1$ by $1$ .

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

Step 1.3.11

Reorder terms.

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

Step 2

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

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

Step 3

Apply L'Hospital's rule.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 3.1.2.1

Evaluate the limit.

Tap for more steps...

Step 3.1.2.1.1

Move the exponent $2$ from ${(x - 5)}^{2}$ outside the limit using the Limits Power Rule.

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

Step 3.1.2.1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

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

Step 3.1.2.1.3

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

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

Step 3.1.2.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

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

Step 3.1.2.3

Simplify the answer.

Tap for more steps...

Step 3.1.2.3.1

Multiply $- 1$ by $5$ .

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

Step 3.1.2.3.2

Subtract $5$ from $5$ .

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

Step 3.1.2.3.3

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

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

Step 3.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 3.1.3.1

Evaluate the limit.

Tap for more steps...

Step 3.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

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

Step 3.1.3.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $5$ .

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

Step 3.1.3.1.3

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

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

Step 3.1.3.1.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

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

Step 3.1.3.1.5

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

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

Step 3.1.3.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$3 \frac{0}{- 1 + \cos (5 - 1 \cdot 5)}$

Step 3.1.3.3

Simplify the answer.

Tap for more steps...

Step 3.1.3.3.1

Simplify each term.

Tap for more steps...

Step 3.1.3.3.1.1

Multiply $- 1$ by $5$ .

$3 \frac{0}{- 1 + \cos (5 - 5)}$

Step 3.1.3.3.1.2

Subtract $5$ from $5$ .

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

Step 3.1.3.3.1.3

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

$3 \frac{0}{- 1 + 1}$

Step 3.1.3.3.2

Add $- 1$ and $1$ .

$3 (\frac{0}{0})$

Step 3.1.3.3.3

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

Undefined

$3 (\frac{0}{0})$

Step 3.1.3.4

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

Undefined

$3 (\frac{0}{0})$

Step 3.1.4

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

Undefined

$3 (\frac{0}{0})$

Step 3.2

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.

$lim_{x \to 5} \frac{{(x - 5)}^{2}}{- 1 + \cos (x - 5)} = lim_{x \to 5} \frac{\frac{d}{d x} [{(x - 5)}^{2}]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

Rewrite ${(x - 5)}^{2}$ as $(x - 5) (x - 5)$ .

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

Step 3.3.3

Expand $(x - 5) (x - 5)$ using the FOIL Method.

Tap for more steps...

Step 3.3.3.1

Apply the distributive property.

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

Step 3.3.3.2

Apply the distributive property.

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

Step 3.3.3.3

Apply the distributive property.

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

Step 3.3.4

Simplify and combine like terms.

Tap for more steps...

Step 3.3.4.1

Simplify each term.

Tap for more steps...

Step 3.3.4.1.1

Multiply $x$ by $x$ .

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

Step 3.3.4.1.2

Move $- 5$ to the left of $x$ .

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

Step 3.3.4.1.3

Multiply $- 5$ by $- 5$ .

$3 lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 5 x - 5 x + 25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.4.2

Subtract $5 x$ from $- 5 x$ .

$3 lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 10 x + 25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.5

By the Sum Rule, the derivative of $x^{2} - 10 x + 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [25]$ .

$3 lim_{x \to 5} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.6

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

$3 lim_{x \to 5} \frac{2 x + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.7

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x$ with respect to $x$ is $- 10 \frac{d}{d x} [x]$ .

$3 lim_{x \to 5} \frac{2 x - 10 \frac{d}{d x} [x] + \frac{d}{d x} [25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.8

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

$3 lim_{x \to 5} \frac{2 x - 10 \cdot 1 + \frac{d}{d x} [25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.9

Multiply $- 10$ by $1$ .

$3 lim_{x \to 5} \frac{2 x - 10 + \frac{d}{d x} [25]}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.10

Since $25$ is constant with respect to $x$ , the derivative of $25$ with respect to $x$ is $0$ .

$3 lim_{x \to 5} \frac{2 x - 10 + 0}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.11

Add $2 x - 10$ and $0$ .

$3 lim_{x \to 5} \frac{2 x - 10}{\frac{d}{d x} [- 1 + \cos (x - 5)]}$

Step 3.3.12

By the Sum Rule, the derivative of $- 1 + \cos (x - 5)$ with respect to $x$ is $\frac{d}{d x} [- 1] + \frac{d}{d x} [\cos (x - 5)]$ .

$3 lim_{x \to 5} \frac{2 x - 10}{\frac{d}{d x} [- 1] + \frac{d}{d x} [\cos (x - 5)]}$

Step 3.3.13

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

$3 lim_{x \to 5} \frac{2 x - 10}{0 + \frac{d}{d x} [\cos (x - 5)]}$

Step 3.3.14

Evaluate $\frac{d}{d x} [\cos (x - 5)]$ .

Tap for more steps...

Step 3.3.14.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = x - 5$ .

Tap for more steps...

Step 3.3.14.1.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [x - 5]}$

Step 3.3.14.1.2

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

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (u) \frac{d}{d x} [x - 5]}$

Step 3.3.14.1.3

Replace all occurrences of $u$ with $x - 5$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5) \frac{d}{d x} [x - 5]}$

Step 3.3.14.2

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 5])}$

Step 3.3.14.3

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

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5) (1 + \frac{d}{d x} [- 5])}$

Step 3.3.14.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5) (1 + 0)}$

Step 3.3.14.5

Add $1$ and $0$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5) \cdot 1}$

Step 3.3.14.6

Multiply $- 1$ by $1$ .

$3 lim_{x \to 5} \frac{2 x - 10}{0 - \sin (x - 5)}$

Step 3.3.15

Subtract $\sin (x - 5)$ from $0$ .

$3 lim_{x \to 5} \frac{2 x - 10}{- \sin (x - 5)}$

Step 4

Apply L'Hospital's rule.

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

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

$3 \frac{lim_{x \to 5} 2 x - 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 4.1.2.1

Evaluate the limit.

Tap for more steps...

Step 4.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$3 \frac{lim_{x \to 5} 2 x - lim_{x \to 5} 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.1.2

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

$3 \frac{2 lim_{x \to 5} x - lim_{x \to 5} 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.1.3

Evaluate the limit of $10$ which is constant as $x$ approaches $5$ .

$3 \frac{2 lim_{x \to 5} x - 1 \cdot 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$3 \frac{2 \cdot 5 - 1 \cdot 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.3

Simplify the answer.

Tap for more steps...

Step 4.1.2.3.1

Simplify each term.

Tap for more steps...

Step 4.1.2.3.1.1

Multiply $2$ by $5$ .

$3 \frac{10 - 1 \cdot 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.3.1.2

Multiply $- 1$ by $10$ .

$3 \frac{10 - 10}{lim_{x \to 5} - \sin (x - 5)}$

Step 4.1.2.3.2

Subtract $10$ from $10$ .

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

Step 4.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 4.1.3.1

Evaluate the limit.

Tap for more steps...

Step 4.1.3.1.1

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

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

Step 4.1.3.1.2

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

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

Step 4.1.3.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

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

Step 4.1.3.1.4

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

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

Step 4.1.3.2

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$3 \frac{0}{- \sin (5 - 1 \cdot 5)}$

Step 4.1.3.3

Simplify the answer.

Tap for more steps...

Step 4.1.3.3.1

Multiply $- 1$ by $5$ .

$3 \frac{0}{- \sin (5 - 5)}$

Step 4.1.3.3.2

Subtract $5$ from $5$ .

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

Step 4.1.3.3.3

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

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

Step 4.1.3.3.4

Multiply $- 1$ by $0$ .

$3 (\frac{0}{0})$

Step 4.1.3.3.5

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

Undefined

$3 (\frac{0}{0})$

Step 4.1.3.4

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

Undefined

$3 (\frac{0}{0})$

Step 4.1.4

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

Undefined

$3 (\frac{0}{0})$

Step 4.2

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.

$lim_{x \to 5} \frac{2 x - 10}{- \sin (x - 5)} = lim_{x \to 5} \frac{\frac{d}{d x} [2 x - 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 4.3.1

Differentiate the numerator and denominator.

$3 lim_{x \to 5} \frac{\frac{d}{d x} [2 x - 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3.2

By the Sum Rule, the derivative of $2 x - 10$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 10]$ .

$3 lim_{x \to 5} \frac{\frac{d}{d x} [2 x] + \frac{d}{d x} [- 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3.3

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

Tap for more steps...

Step 4.3.3.1

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

$3 lim_{x \to 5} \frac{2 \frac{d}{d x} [x] + \frac{d}{d x} [- 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3.3.2

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

$3 lim_{x \to 5} \frac{2 \cdot 1 + \frac{d}{d x} [- 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3.3.3

Multiply $2$ by $1$ .

$3 lim_{x \to 5} \frac{2 + \frac{d}{d x} [- 10]}{\frac{d}{d x} [- \sin (x - 5)]}$

Step 4.3.4

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

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

Step 4.3.5

Add $2$ and $0$ .

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

Step 4.3.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x - 5)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x - 5)]$ .

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

Step 4.3.7

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 $g (x) = x - 5$ .

Tap for more steps...

Step 4.3.7.1

To apply the Chain Rule, set $u$ as $x - 5$ .

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

Step 4.3.7.2

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

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

Step 4.3.7.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 4.3.8

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

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

Step 4.3.9

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

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

Step 4.3.10

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

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

Step 4.3.11

Add $1$ and $0$ .

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

Step 4.3.12

Multiply $- 1$ by $1$ .

$3 lim_{x \to 5} \frac{2}{- \cos (x - 5)}$

Step 5

Evaluate the limit.

Tap for more steps...

Step 5.1

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

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

Step 5.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $5$ .

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

Step 5.3

Evaluate the limit of $1$ which is constant as $x$ approaches $5$ .

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

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

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

Step 5.7

Evaluate the limit of $5$ which is constant as $x$ approaches $5$ .

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

Step 6

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

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

Step 7

Simplify the answer.

Tap for more steps...

Step 7.1

Multiply $3$ by $2$ .

$6 \frac{1}{- \cos (5 - 1 \cdot 5)}$

Step 7.2

Cancel the common factor of $1$ and $- 1$ .

Tap for more steps...

Step 7.2.1

Rewrite $1$ as $- 1 (- 1)$ .

$6 \frac{- 1 (- 1)}{- \cos (5 - 1 \cdot 5)}$

Step 7.2.2

Move the negative in front of the fraction.

$6 (- \frac{1}{\cos (5 - 1 \cdot 5)})$

Step 7.3

Convert from $\frac{1}{\cos (5 - 1 \cdot 5)}$ to $\sec (5 - 1 \cdot 5)$ .

$6 (- \sec (5 - 1 \cdot 5))$

Step 7.4

Multiply $- 1$ by $5$ .

$6 (- \sec (5 - 5))$

Step 7.5

Subtract $5$ from $5$ .

$6 (- \sec (0))$

Step 7.6

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

$6 (- 1 \cdot 1)$

Step 7.7

Multiply $6 (- 1 \cdot 1)$ .

Tap for more steps...

Step 7.7.1

Multiply $- 1$ by $1$ .

$6 \cdot - 1$

Step 7.7.2

Multiply $6$ by $- 1$ .

$- 6$