# Find the limit. lim_{x to 5} {5 - x} / {x^2 - 25}.

## Question:

Find the limit. {eq}\displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} {/eq}.

## Limits:

The limit is the process in which the function f(x) approaches to some value as x approaches to some value. The limit of the function can be evaluated by using the algebraic operations.

## Answer and Explanation:

We have:

{eq}\displaystyle \lim_{x \to 5} \dfrac{5 - x}{x^2 - 25} {/eq}

At x = 5

{eq}\dfrac {5 - 5} {5^2 - 25} = \dfrac{0}{0} {/eq}

The given limit is in the indeterminate form.

Reducing the given limit by using the algebraic operations.

{eq}\displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} {/eq}

Using:

{eq}a^2 - b^2 = (a + b)(a - b) {/eq}

We get:

{eq}\displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} = \lim_{x \to 5} \dfrac {5 - x}{(x + 5)(x - 5)} \\ \displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} = \lim_{x \to 5} \dfrac {5 - x}{-(x + 5)(5 - x)} {/eq}

Dividing the numerator and denominator by 5 - x, we get:

{eq}\displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} = \lim_{x \to 5} \dfrac{1}{-(x + 5)} {/eq}

Substituting the value of limit, we get:

{eq}\displaystyle \lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} = -\dfrac{1}{5 + 5} \\ \displaystyle \boxed{\lim_{x \to 5} \dfrac {5 - x} {x^2 - 25} = -\dfrac{1}{10}} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question