Simplify limit as x approaches 5 of (x^2-25)/(x-5)

$lim_{x \to 5} \frac{x^{2} - 25}{x - 5}$

Step 1

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.

$\frac{lim_{x \to 5} x^{2} - 25}{lim_{x \to 5} x - 5}$

Evaluate the limit of the numerator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{lim_{x \to 5} x^{2} - lim_{x \to 5} 25}{lim_{x \to 5} x - 5}$

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

$\frac{{(lim_{x \to 5} x)}^{2} - lim_{x \to 5} 25}{lim_{x \to 5} x - 5}$

Evaluate the limit of $25$ which is constant as $x$ approaches $5$ .

$\frac{{(lim_{x \to 5} x)}^{2} - 1 \cdot 25}{lim_{x \to 5} x - 5}$

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{5^{2} - 1 \cdot 25}{lim_{x \to 5} x - 5}$

Simplify the answer.

Simplify each term.

Raise $5$ to the power of $2$ .

$\frac{25 - 1 \cdot 25}{lim_{x \to 5} x - 5}$

Multiply $- 1$ by $25$ .

$\frac{25 - 25}{lim_{x \to 5} x - 5}$

Subtract $25$ from $25$ .

$\frac{0}{lim_{x \to 5} x - 5}$

Evaluate the limit of the denominator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $5$ .

$\frac{0}{lim_{x \to 5} x - lim_{x \to 5} 5}$

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

$\frac{0}{lim_{x \to 5} x - 1 \cdot 5}$

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$\frac{0}{5 - 1 \cdot 5}$

Simplify the answer.

Multiply $- 1$ by $5$ .

$\frac{0}{5 - 5}$

Subtract $5$ from $5$ .

$\frac{0}{0}$

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.

$lim_{x \to 5} \frac{x^{2} - 25}{x - 5} = lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 25]}{\frac{d}{d x} [x - 5]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2} - 25]}{\frac{d}{d x} [x - 5]}$

By the Sum Rule, the derivative of $x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]$ .

$lim_{x \to 5} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]}{\frac{d}{d x} [x - 5]}$

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

$lim_{x \to 5} \frac{2 x + \frac{d}{d x} [- 25]}{\frac{d}{d x} [x - 5]}$

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$lim_{x \to 5} \frac{2 x + 0}{\frac{d}{d x} [x - 5]}$

Add $2 x$ and $0$ .

$lim_{x \to 5} \frac{2 x}{\frac{d}{d x} [x - 5]}$

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

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

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

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

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

$lim_{x \to 5} \frac{2 x}{1 + 0}$

Add $1$ and $0$ .

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

Divide $2 x$ by $1$ .

$lim_{x \to 5} 2 x$

Step 2

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

$2 lim_{x \to 5} x$

Step 3

Evaluate the limit of $x$ by plugging in $5$ for $x$ .

$2 \cdot 5$

Step 4

Multiply $2$ by $5$ .

$10$

Simplify limit as x approaches 5 of (x^2-25)/(x-5)

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