Calculus Examples

$lim_{x \to - 1} \frac{2^{x + 1} - 1}{\sin (π x)}$

Step 1

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

Tap for more steps...

Step 1.1

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

$\frac{lim_{x \to - 1} 2^{x + 1} - 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.2.1

Evaluate the limit.

Tap for more steps...

Step 1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{lim_{x \to - 1} 2^{x + 1} - lim_{x \to - 1} 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.1.2

Move the limit into the exponent.

$\frac{2^{lim_{x \to - 1} x + 1} - lim_{x \to - 1} 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{2^{lim_{x \to - 1} x + lim_{x \to - 1} 1} - lim_{x \to - 1} 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.1.4

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

$\frac{2^{lim_{x \to - 1} x + 1} - lim_{x \to - 1} 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.1.5

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

$\frac{2^{lim_{x \to - 1} x + 1} - 1 \cdot 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{2^{- 1 + 1} - 1 \cdot 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.3

Simplify the answer.

Tap for more steps...

Step 1.2.3.1

Simplify each term.

Tap for more steps...

Step 1.2.3.1.1

Add $- 1$ and $1$ .

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

Step 1.2.3.1.2

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.3.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to - 1} \sin (π x)}$

Step 1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to - 1} \sin (π x)}$

Step 1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.3.1

Evaluate the limit.

Tap for more steps...

Step 1.3.1.1

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

$\frac{0}{\sin (lim_{x \to - 1} π x)}$

Step 1.3.1.2

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

$\frac{0}{\sin (π lim_{x \to - 1} x)}$

Step 1.3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{0}{\sin (π \cdot - 1)}$

Step 1.3.3

Simplify the answer.

Tap for more steps...

Step 1.3.3.1

Move $- 1$ to the left of $π$ .

$\frac{0}{\sin (- 1 \cdot π)}$

Step 1.3.3.2

Rewrite $- 1 π$ as $- π$ .

$\frac{0}{\sin (- π)}$

Step 1.3.3.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 1.3.3.4

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

$\frac{0}{0}$

Step 1.3.3.5

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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.

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

Step 3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

By the Sum Rule, the derivative of $2^{x + 1} - 1$ with respect to $x$ is $\frac{d}{d x} [2^{x + 1}] + \frac{d}{d x} [- 1]$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [2^{x + 1}]$ .

Tap for more steps...

Step 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 $f (x) = 2^{x}$ and $g (x) = x + 1$ .

Tap for more steps...

Step 3.3.1.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 3.3.1.2

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

$lim_{x \to - 1} \frac{2^{u} \ln (2) \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [\sin (π x)]}$

Step 3.3.1.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{2^{x + 1} \ln (2) \cdot 1 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [\sin (π x)]}$

Step 3.3.6

Multiply $2^{x + 1}$ by $1$ .

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

Step 3.4

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

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

Step 3.5

Add $2^{x + 1} \ln (2)$ and $0$ .

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

Step 3.6

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 $g (x) = π x$ .

Tap for more steps...

Step 3.6.1

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

$lim_{x \to - 1} \frac{2^{x + 1} \ln (2)}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [π x]}$

Step 3.6.2

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

$lim_{x \to - 1} \frac{2^{x + 1} \ln (2)}{\cos (u) \frac{d}{d x} [π x]}$

Step 3.6.3

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

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Multiply $π$ by $1$ .

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

Step 3.10

Reorder the factors of $\cos (π x) π$ .

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

Step 4

Move the term $\frac{\ln (2)}{π}$ outside of the limit because it is constant with respect to $x$ .

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

Step 5

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 1$ .

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

Step 6

Move the limit into the exponent.

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

Step 7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

Tap for more steps...

Step 11.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 11.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

$\frac{\ln (2)}{π} \cdot \frac{2^{- 1 + 1}}{\cos (π \cdot - 1)}$

Step 12

Simplify the answer.

Tap for more steps...

Step 12.1

Combine.

$\frac{\ln (2) \cdot 2^{- 1 + 1}}{π \cos (π \cdot - 1)}$

Step 12.2

Simplify the numerator.

Tap for more steps...

Step 12.2.1

Add $- 1$ and $1$ .

$\frac{\ln (2) \cdot 2^{0}}{π \cos (π \cdot - 1)}$

Step 12.2.2

Anything raised to $0$ is $1$ .

$\frac{\ln (2) \cdot 1}{π \cos (π \cdot - 1)}$

Step 12.3

Simplify the denominator.

Tap for more steps...

Step 12.3.1

Move $- 1$ to the left of $π$ .

$\frac{\ln (2) \cdot 1}{π \cos (- 1 \cdot π)}$

Step 12.3.2

Rewrite $- 1 π$ as $- π$ .

$\frac{\ln (2) \cdot 1}{π \cos (- π)}$

Step 12.3.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$\frac{\ln (2) \cdot 1}{π (- \cos (0))}$

Step 12.3.4

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

$\frac{\ln (2) \cdot 1}{π (- 1 \cdot 1)}$

Step 12.3.5

Multiply $- 1$ by $1$ .

$\frac{\ln (2) \cdot 1}{π \cdot - 1}$

Step 12.4

Multiply $\ln (2)$ by $1$ .

$\frac{\ln (2)}{π \cdot - 1}$

Step 12.5

Move $- 1$ to the left of $π$ .

$\frac{\ln (2)}{- 1 \cdot π}$

Step 12.6

Move the negative in front of the fraction.

$- \frac{\ln (2)}{π}$