Calculus Examples

$lim_{x \to 1} \frac{x^{2} + 4 x - 5}{x^{2} - 1}$

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.

$\frac{lim_{x \to 1} x^{2} + 4 x - 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 1} x^{2} + lim_{x \to 1} 4 x - lim_{x \to 1} 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to 1} x)}^{2} + lim_{x \to 1} 4 x - lim_{x \to 1} 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to 1} x)}^{2} + 4 lim_{x \to 1} x - lim_{x \to 1} 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.4

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

$\frac{{(lim_{x \to 1} x)}^{2} + 4 lim_{x \to 1} x - 1 \cdot 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.5

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

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Step 1.1.2.5.1

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

$\frac{1^{2} + 4 lim_{x \to 1} x - 1 \cdot 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.5.2

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

$\frac{1^{2} + 4 \cdot 1 - 1 \cdot 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.6

Simplify the answer.

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Step 1.1.2.6.1

Simplify each term.

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Step 1.1.2.6.1.1

One to any power is one.

$\frac{1 + 4 \cdot 1 - 1 \cdot 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.6.1.2

Multiply $4$ by $1$ .

$\frac{1 + 4 - 1 \cdot 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.6.1.3

Multiply $- 1$ by $5$ .

$\frac{1 + 4 - 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.6.2

Add $1$ and $4$ .

$\frac{5 - 5}{lim_{x \to 1} x^{2} - 1}$

Step 1.1.2.6.3

Subtract $5$ from $5$ .

$\frac{0}{lim_{x \to 1} x^{2} - 1}$

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 $1$ .

$\frac{0}{lim_{x \to 1} x^{2} - lim_{x \to 1} 1}$

Step 1.1.3.1.2

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

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

$\frac{0}{1^{2} - 1 \cdot 1}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

One to any power is one.

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

Step 1.1.3.3.1.2

Multiply $- 1$ by $1$ .

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

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 1} \frac{x^{2} + 4 x - 5}{x^{2} - 1} = lim_{x \to 1} \frac{\frac{d}{d x} [x^{2} + 4 x - 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 1} \frac{\frac{d}{d x} [x^{2} + 4 x - 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} + 4 x - 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 5]$ .

$lim_{x \to 1} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.3

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

$lim_{x \to 1} \frac{2 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 1.3.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to 1} \frac{2 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.4.2

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

$lim_{x \to 1} \frac{2 x + 4 \cdot 1 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.4.3

Multiply $4$ by $1$ .

$lim_{x \to 1} \frac{2 x + 4 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x^{2} - 1]}$

Step 1.3.5

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

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

Step 1.3.6

Add $2 x + 4$ and $0$ .

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

Step 1.3.7

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

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

Step 1.3.8

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

$lim_{x \to 1} \frac{2 x + 4}{2 x + \frac{d}{d x} [- 1]}$

Step 1.3.9

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

$lim_{x \to 1} \frac{2 x + 4}{2 x + 0}$

Step 1.3.10

Add $2 x$ and $0$ .

$lim_{x \to 1} \frac{2 x + 4}{2 x}$

Step 1.4

Cancel the common factor of $2 x + 4$ and $2$ .

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Step 1.4.1

Factor $2$ out of $2 x$ .

$lim_{x \to 1} \frac{2 (x) + 4}{2 x}$

Step 1.4.2

Factor $2$ out of $4$ .

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

Step 1.4.3

Factor $2$ out of $2 (x) + 2 (2)$ .

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

Step 1.4.4

Cancel the common factors.

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Step 1.4.4.1

Factor $2$ out of $2 x$ .

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

Step 1.4.4.2

Cancel the common factor.

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

Step 1.4.4.3

Rewrite the expression.

$lim_{x \to 1} \frac{x + 2}{x}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{lim_{x \to 1} x + 2}{lim_{x \to 1} x}$

Step 2.2

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

$\frac{lim_{x \to 1} x + lim_{x \to 1} 2}{lim_{x \to 1} x}$

Step 2.3

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$\frac{lim_{x \to 1} x + 2}{lim_{x \to 1} x}$

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 $x$ .

$\frac{1 + 2}{lim_{x \to 1} x}$

Step 3.2

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

$\frac{1 + 2}{1}$

Step 4

Simplify the answer.

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Step 4.1

Divide $1 + 2$ by $1$ .

$1 + 2$

Step 4.2

Add $1$ and $2$ .

$3$