lim_(x rarr5)(x^(2)-25)/(x-5)=_____ (A) 7 (B) 5 (C) 10 (D) 9 (E) 0

Identify indeterminate form: Identify the indeterminate form. We need to evaluate the limit of the function (x^2 - 25)/(x - 5) as x approaches 5. If we substitute x = 5 directly into the function, we get (5^2 - 25)/(5 - 5) = 0/0, which is an indeterminate form.Factor numerator: Factor the numerator. The numerator x^2 - 25 is a difference of squares and can be factored as (x + 5)(x - 5).Simplify expression: Simplify the expression. We can simplify the expression by canceling out the common factor (x - 5) from the numerator and the denominator. So, (x^2 - 25)/(x - 5) becomes (x + 5)(x - 5)/(x - 5) = x + 5, when x ≠ 5.Evaluate limit: Evaluate the limit. Now that we have simplified the expression, we can find the limit as x approaches 5 by substituting x = 5 into the simplified expression. lim_(x -> 5)(x + 5) = 5 + 5 = 10.

Home

Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=\_\_\_\_\_

\newline

(A)

7

\newline

(B)

5

\newline

(C)

10

\newline

(D)

9

\newline

(E)

0

\newline

Identify indeterminate form: Identify the indeterminate form. $\newline$ We need to evaluate the limit of the function $(x^2 - 25)/(x - 5)$ as $x$ approaches $5$ . If we substitute $x = 5$ directly into the function, we get $(5^2 - 25)/(5 - 5) = 0/0$ , which is an indeterminate form.
Factor numerator: Factor the numerator. $\newline$ The numerator $x^2 - 25$ is a difference of squares and can be factored as $(x + 5)(x - 5)$ .
Simplify expression: Simplify the expression. $\newline$ We can simplify the expression by canceling out the common factor $(x - 5)$ from the numerator and the denominator. $\newline$ So, $(x^2 - 25)/(x - 5)$ becomes $(x + 5)(x - 5)/(x - 5) = x + 5$ , when $x \neq 5$ .
Evaluate limit: Evaluate the limit. $\newline$ Now that we have simplified the expression, we can find the limit as $x$ approaches $5$ by substituting $x = 5$ into the simplified expression. $\newline$ $\lim_{x \to 5}(x + 5) = 5 + 5 = 10$ .