Find lim_(x rarr-4)(x^(2)-16)/(x^(2)+4x). Choose 1 answer: (A) 2 (B) -2 (c) 0 (D) The limit doesn't exist

Substitute and Check: First, let's try to directly substitute the value of x with -4 into the function to see if it results in an indeterminate form. Substitute -4 for x in (x^2 - 16)/(x^2 + 4x). lim_(x rarr-4)(x^2 - 16)/(x^2 + 4x) = ((-4)^2 - 16)/((-4)^2 + 4(-4)) = (16 - 16)/(16 - 16) = 0/0 We get an indeterminate form of 0/0, which means we need to simplify the expression further to find the limit.Factor Numerator and Denominator: Since we have an indeterminate form, we can try to factor the numerator and the denominator to see if any common factors can be canceled out. Factor the numerator (x^2 - 16) and the denominator (x^2 + 4x). The numerator is a difference of squares and can be factored as (x + 4)(x - 4). The denominator can be factored by taking x common, resulting in x(x + 4). So the expression becomes: lim_(x rarr-4)((x + 4)(x - 4))/(x(x + 4))Cancel Common Factor: Now, we can cancel out the common factor (x + 4) from the numerator and the denominator. lim_(x rarr-4)((x + 4)(x - 4))/(x(x + 4)) = lim_(x rarr-4)(x - 4)/x Now that the expression is simplified, we can substitute -4 for x again to find the limit.Substitute and Simplify: Substitute -4 for x in the simplified expression (x - 4)/x. lim_(x rarr-4)(x - 4)/x = (-4 - 4)/(-4) = (-8)/(-4) = 2 The limit exists and is equal to 2.

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Find
lim_(x rarr-4)(x^(2)-16)/(x^(2)+4x).
Choose 1 answer:
(A) 2
(B) -2
(c) 0
(D) The limit doesn't exist

Find $\lim _{x \rightarrow-4} \frac{x^{2}-16}{x^{2}+4 x}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $2$ $\newline$ (B) $-2$ $\newline$ (C) $0$ $\newline$ (D) The limit doesn't exist

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Math Problems

Algebra 2

Evaluate exponential functions

Full solution

Q. Find

\lim _{x \rightarrow-4} \frac{x^{2}-16}{x^{2}+4 x}

\newline

Choose

1

answer:

\newline

(A)

2

\newline

(B)

-2

\newline

(C)

0

\newline

(D) The limit doesn't exist

Substitute and Check: First, let's try to directly substitute the value of $x$ with $-4$ into the function to see if it results in an indeterminate form. $\newline$ Substitute $-4$ for $x$ in $(x^2 - 16)/(x^2 + 4x)$ . $\newline$ $\lim_{x \to -4}(x^2 - 16)/(x^2 + 4x)$ $\newline$ = ((\-4)^2 - 16)/((\-4)^2 + 4(\-4)) $\newline$ = $(16 - 16)/(16 - 16)$ $\newline$ = $0/0$ $\newline$ We get an indeterminate form of $0/0$ , which means we need to simplify the expression further to find the limit.
Factor Numerator and Denominator: Since we have an indeterminate form, we can try to factor the numerator and the denominator to see if any common factors can be canceled out. $\newline$ Factor the numerator $(x^2 - 16)$ and the denominator $(x^2 + 4x)$ . $\newline$ The numerator is a difference of squares and can be factored as $(x + 4)(x - 4)$ . $\newline$ The denominator can be factored by taking $x$ common, resulting in $x(x + 4)$ . $\newline$ So the expression becomes: $\newline$ $\lim_{x \rightarrow -4}\frac{(x + 4)(x - 4)}{x(x + 4)}$
Cancel Common Factor: Now, we can cancel out the common factor $(x + 4)$ from the numerator and the denominator. $\newline$ $\lim_{x \to -4}\frac{(x + 4)(x - 4)}{x(x + 4)} = \lim_{x \to -4}\frac{x - 4}{x}$ $\newline$ Now that the expression is simplified, we can substitute $-4$ for $x$ again to find the limit.
Substitute and Simplify: Substitute $-4$ for $x$ in the simplified expression $(x - 4)/x$ . $\newline$ $\lim_{x \to -4}\frac{x - 4}{x}$ $\newline$ = \frac{ $-4$ - $4$ }{ $-4$ } $\newline$ = \frac{ $-8$ }{ $-4$ } $\newline$ = $2$ $\newline$ The limit exists and is equal to $2$ .