Calculus Examples

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

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.

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

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

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

Tap for more steps...

Step 1.1.2.3.1

Simplify each term.

Tap for more steps...

Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $- 1$ by $16$ .

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

Step 1.1.2.3.2

Subtract $16$ from $16$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

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

Step 1.1.3.4

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

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

Step 1.1.3.5

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

Tap for more steps...

Step 1.1.3.5.1

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

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

Step 1.1.3.5.2

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

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

Step 1.1.3.6

Simplify the answer.

Tap for more steps...

Step 1.1.3.6.1

Simplify each term.

Tap for more steps...

Step 1.1.3.6.1.1

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

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

Step 1.1.3.6.1.2

Multiply $- 8$ by $4$ .

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

Step 1.1.3.6.2

Subtract $32$ from $16$ .

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

Step 1.1.3.6.3

Add $- 16$ and $16$ .

$\frac{0}{0}$

Step 1.1.3.6.4

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

Undefined

$\frac{0}{0}$

Step 1.1.3.7

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

Tap for more steps...

Step 1.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

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 $n = 2$ .

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

Step 1.3.4.3

Multiply $2$ by $- 1$ .

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

Step 1.3.5

Subtract $2 x$ from $0$ .

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

Step 1.3.6

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

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

Step 1.3.7

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

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

Step 1.3.8

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

Tap for more steps...

Step 1.3.8.1

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

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

Step 1.3.8.2

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

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

Step 1.3.8.3

Multiply $- 8$ by $1$ .

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

Step 1.3.9

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

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

Step 1.3.10

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

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

Step 1.4

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

Tap for more steps...

Step 1.4.1

Factor $2$ out of $- 2 x$ .

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

Step 1.4.2

Cancel the common factors.

Tap for more steps...

Step 1.4.2.1

Factor $2$ out of $2 x$ .

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

Step 1.4.2.2

Factor $2$ out of $- 8$ .

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

Step 1.4.2.3

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

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

Step 1.4.2.4

Cancel the common factor.

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

Step 1.4.2.5

Rewrite the expression.

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

Step 2

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

$Does not exist$