Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

Step 1.1.2.4

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

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

Step 1.1.2.5

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

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

Step 1.1.2.6

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

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

Step 1.1.2.7

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

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Step 1.1.2.7.1

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

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

Step 1.1.2.7.2

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

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

Step 1.1.2.8

Simplify the answer.

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Step 1.1.2.8.1

Simplify each term.

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Step 1.1.2.8.1.1

Raise $- 1$ to the power of $2$ .

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

Step 1.1.2.8.1.2

Multiply $2$ by $1$ .

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

Step 1.1.2.8.1.3

Multiply $3$ by $- 1$ .

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

Step 1.1.2.8.2

Subtract $3$ from $2$ .

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

Step 1.1.2.8.3

Add $- 1$ and $1$ .

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

Step 1.1.2.8.4

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

$\frac{0}{lim_{x \to - 1} x + 1}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

$\frac{0}{lim_{x \to - 1} x + lim_{x \to - 1} 1}$

Step 1.1.3.1.2

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

$\frac{0}{lim_{x \to - 1} x + 1}$

Step 1.1.3.2

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

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

Step 1.1.3.3

Add $- 1$ and $1$ .

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

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 $f (x) = \sin (x)$ and $g (x) = 2 x^{2} + 3 x + 1$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $2 x^{2} + 3 x + 1$ .

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

Step 1.3.2.2

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

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

Step 1.3.2.3

Replace all occurrences of $u$ with $2 x^{2} + 3 x + 1$ .

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

Step 1.3.3

By the Sum Rule, the derivative of $2 x^{2} + 3 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $2$ by $2$ .

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.3.9

Multiply $3$ by $1$ .

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

Step 1.3.10

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

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

Step 1.3.11

Add $4 x + 3$ and $0$ .

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

Step 1.3.12

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

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

Step 1.3.13

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

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

Step 1.3.14

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

$lim_{x \to - 1} \frac{\cos (2 x^{2} + 3 x + 1) (4 x + 3)}{1 + 0}$

Step 1.3.15

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{\cos (2 x^{2} + 3 x + 1) (4 x + 3)}{1}$

Step 1.4

Divide $\cos (2 x^{2} + 3 x + 1) (4 x + 3)$ by $1$ .

$lim_{x \to - 1} \cos (2 x^{2} + 3 x + 1) (4 x + 3)$

Step 2

Evaluate the limit.

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Step 2.1

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

$lim_{x \to - 1} \cos (2 x^{2} + 3 x + 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.2

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

$\cos (lim_{x \to - 1} 2 x^{2} + 3 x + 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.3

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

$\cos (lim_{x \to - 1} 2 x^{2} + lim_{x \to - 1} 3 x + lim_{x \to - 1} 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.4

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

$\cos (2 lim_{x \to - 1} x^{2} + lim_{x \to - 1} 3 x + lim_{x \to - 1} 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.5

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

$\cos (2 {(lim_{x \to - 1} x)}^{2} + lim_{x \to - 1} 3 x + lim_{x \to - 1} 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.6

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

$\cos (2 {(lim_{x \to - 1} x)}^{2} + 3 lim_{x \to - 1} x + lim_{x \to - 1} 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.7

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

$\cos (2 {(lim_{x \to - 1} x)}^{2} + 3 lim_{x \to - 1} x + 1) \cdot lim_{x \to - 1} 4 x + 3$

Step 2.8

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

$\cos (2 {(lim_{x \to - 1} x)}^{2} + 3 lim_{x \to - 1} x + 1) \cdot (lim_{x \to - 1} 4 x + lim_{x \to - 1} 3)$

Step 2.9

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

$\cos (2 {(lim_{x \to - 1} x)}^{2} + 3 lim_{x \to - 1} x + 1) \cdot (4 lim_{x \to - 1} x + lim_{x \to - 1} 3)$

Step 2.10

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

$\cos (2 {(lim_{x \to - 1} x)}^{2} + 3 lim_{x \to - 1} x + 1) \cdot (4 lim_{x \to - 1} x + 3)$

Step 3

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

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Step 3.1

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

$\cos (2 {(- 1)}^{2} + 3 lim_{x \to - 1} x + 1) \cdot (4 lim_{x \to - 1} x + 3)$

Step 3.2

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

$\cos (2 {(- 1)}^{2} + 3 \cdot - 1 + 1) \cdot (4 lim_{x \to - 1} x + 3)$

Step 3.3

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

$\cos (2 {(- 1)}^{2} + 3 \cdot - 1 + 1) \cdot (4 \cdot - 1 + 3)$

Step 4

Simplify the answer.

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $- 1$ to the power of $2$ .

$\cos (2 \cdot 1 + 3 \cdot - 1 + 1) \cdot (4 \cdot - 1 + 3)$

Step 4.1.2

Multiply $2$ by $1$ .

$\cos (2 + 3 \cdot - 1 + 1) \cdot (4 \cdot - 1 + 3)$

Step 4.1.3

Multiply $3$ by $- 1$ .

$\cos (2 - 3 + 1) \cdot (4 \cdot - 1 + 3)$

Step 4.2

Subtract $3$ from $2$ .

$\cos (- 1 + 1) \cdot (4 \cdot - 1 + 3)$

Step 4.3

Add $- 1$ and $1$ .

$\cos (0) \cdot (4 \cdot - 1 + 3)$

Step 4.4

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

$1 \cdot (4 \cdot - 1 + 3)$

Step 4.5

Multiply $4 \cdot - 1 + 3$ by $1$ .

$4 \cdot - 1 + 3$

Step 4.6

Multiply $4$ by $- 1$ .

$- 4 + 3$

Step 4.7

Add $- 4$ and $3$ .

$- 1$