Simplify limit as x approaches 4 of (x^2+2x-24)/(x^2-8x+16)

$lim_{x \to 4} \frac{x^{2} + 2 x - 24}{x^{2} - 8 x + 16}$

Apply L\'Hospital\'s rule.

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

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

$\frac{lim_{x \to 4} x^{2} + 2 x - 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Evaluate the limit of the numerator.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{lim_{x \to 4} x^{2} + lim_{x \to 4} 2 x - lim_{x \to 4} 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

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

$\frac{{(lim_{x \to 4} x)}^{2} + lim_{x \to 4} 2 x - lim_{x \to 4} 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

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

$\frac{{(lim_{x \to 4} x)}^{2} + 2 lim_{x \to 4} x - lim_{x \to 4} 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Evaluate the limit of $24$ which is constant as $x$ approaches $4$ .

$\frac{{(lim_{x \to 4} x)}^{2} + 2 lim_{x \to 4} x - 1 \cdot 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Evaluate the limits by plugging in $4$ for all occurrences of $x$ .

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{4^{2} + 2 lim_{x \to 4} x - 1 \cdot 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{4^{2} + 2 \cdot 4 - 1 \cdot 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Simplify the answer.

Simplify each term.

Raise $4$ to the power of $2$ .

$\frac{16 + 2 \cdot 4 - 1 \cdot 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Multiply $2$ by $4$ .

$\frac{16 + 8 - 1 \cdot 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Multiply $- 1$ by $24$ .

$\frac{16 + 8 - 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Add $16$ and $8$ .

$\frac{24 - 24}{lim_{x \to 4} x^{2} - 8 x + 16}$

Subtract $24$ from $24$ .

$\frac{0}{lim_{x \to 4} x^{2} - 8 x + 16}$

Evaluate the limit of the denominator.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{0}{lim_{x \to 4} x^{2} - lim_{x \to 4} 8 x + lim_{x \to 4} 16}$

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

$\frac{0}{{(lim_{x \to 4} x)}^{2} - lim_{x \to 4} 8 x + lim_{x \to 4} 16}$

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{{(lim_{x \to 4} x)}^{2} - 8 lim_{x \to 4} x + lim_{x \to 4} 16}$

Evaluate the limit of $16$ which is constant as $x$ approaches $4$ .

$\frac{0}{{(lim_{x \to 4} x)}^{2} - 8 lim_{x \to 4} x + 16}$

Evaluate the limits by plugging in $4$ for all occurrences of $x$ .

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{0}{4^{2} - 8 lim_{x \to 4} x + 16}$

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{0}{4^{2} - 8 \cdot 4 + 16}$

Simplify the answer.

Simplify each term.

Raise $4$ to the power of $2$ .

$\frac{0}{16 - 8 \cdot 4 + 16}$

Multiply $- 8$ by $4$ .

$\frac{0}{16 - 32 + 16}$

Subtract $32$ from $16$ .

$\frac{0}{- 16 + 16}$

Add $- 16$ and $16$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 4} \frac{x^{2} + 2 x - 24}{x^{2} - 8 x + 16} = lim_{x \to 4} \frac{\frac{d}{d x} [x^{2} + 2 x - 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

$lim_{x \to 4} \frac{\frac{d}{d x} [x^{2} + 2 x - 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

By the Sum Rule, the derivative of $x^{2} + 2 x - 24$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 24]$ .

$lim_{x \to 4} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

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

$lim_{x \to 4} \frac{2 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

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

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

$lim_{x \to 4} \frac{2 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

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

$lim_{x \to 4} \frac{2 x + 2 \cdot 1 + \frac{d}{d x} [- 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

Multiply $2$ by $1$ .

$lim_{x \to 4} \frac{2 x + 2 + \frac{d}{d x} [- 24]}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

Since $- 24$ is constant with respect to $x$ , the derivative of $- 24$ with respect to $x$ is $0$ .

$lim_{x \to 4} \frac{2 x + 2 + 0}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

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

$lim_{x \to 4} \frac{2 x + 2}{\frac{d}{d x} [x^{2} - 8 x + 16]}$

By the Sum Rule, the derivative of $x^{2} - 8 x + 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [16]$ .

$lim_{x \to 4} \frac{2 x + 2}{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [16]}$

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

$lim_{x \to 4} \frac{2 x + 2}{2 x + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [16]}$

Evaluate $\frac{d}{d x} [- 8 x]$ .

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$lim_{x \to 4} \frac{2 x + 2}{2 x - 8 \frac{d}{d x} [x] + \frac{d}{d x} [16]}$

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

$lim_{x \to 4} \frac{2 x + 2}{2 x - 8 \cdot 1 + \frac{d}{d x} [16]}$

Multiply $- 8$ by $1$ .

$lim_{x \to 4} \frac{2 x + 2}{2 x - 8 + \frac{d}{d x} [16]}$

Since $16$ is constant with respect to $x$ , the derivative of $16$ with respect to $x$ is $0$ .

$lim_{x \to 4} \frac{2 x + 2}{2 x - 8 + 0}$

Add $2 x - 8$ and $0$ .

$lim_{x \to 4} \frac{2 x + 2}{2 x - 8}$

Cancel the common factor of $2 x + 2$ and $2 x - 8$ .

Factor $2$ out of $2 x$ .

$lim_{x \to 4} \frac{2 (x) + 2}{2 x - 8}$

Factor $2$ out of $2$ .

$lim_{x \to 4} \frac{2 (x) + 2 \cdot 1}{2 x - 8}$

Factor $2$ out of $2 (x) + 2 (1)$ .

$lim_{x \to 4} \frac{2 (x + 1)}{2 x - 8}$

Cancel the common factors.

Factor $2$ out of $2 x$ .

$lim_{x \to 4} \frac{2 (x + 1)}{2 (x) - 8}$

Factor $2$ out of $- 8$ .

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

Factor $2$ out of $2 x + 2 \cdot - 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Since the function approaches $- \infty$ from the left and $\infty$ from the right, the limit does not exist.

$Does not exist$

Simplify limit as x approaches 4 of (x^2+2x-24)/(x^2-8x+16)

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