Find lim_(x rarr4)(3x)/((x-2)(x+2)). Choose 1 answer: (A) (1)/(4) (B) (3)/(4) (C) 1 (D) The limit doesn't exist

Identify Limit: Identify the limit that needs to be evaluated. We need to find the limit of the function (3x)/((x-2)(x+2)) as x approaches 4.Substitute Value: Substitute the value of x into the function to see if the function is defined at that point. lim_(x -> 4)(3x)/((x-2)(x+2)) = (3*4)/((4-2)(4+2))Perform Calculations: Perform the calculations to evaluate the limit. (3*4)/((4-2)(4+2)) = 12/(2*6) = 12/12 = 1Conclude Limit: Conclude the limit based on the calculations. Since we were able to directly substitute x = 4 into the function and get a result without any indeterminate forms, the limit exists and is equal to 1.

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Find
lim_(x rarr4)(3x)/((x-2)(x+2)).
Choose 1 answer:
(A)
(1)/(4)
(B)
(3)/(4)
(C) 1
(D) The limit doesn't exist

Find $\lim _{x \rightarrow 4} \frac{3 x}{(x-2)(x+2)}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{4}$ $\newline$ (B) $\frac{3}{4}$ $\newline$ (C) $1$ $\newline$ (D) The limit doesn't exist

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Math Problems

Precalculus

Find limits of polynomials and rational functions

Full solution

Q. Find

\lim _{x \rightarrow 4} \frac{3 x}{(x-2)(x+2)}

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{4}

\newline

(B)

\frac{3}{4}

\newline

(C)

1

\newline

(D) The limit doesn't exist

Identify Limit: Identify the limit that needs to be evaluated. $\newline$ We need to find the limit of the function $\frac{3x}{(x-2)(x+2)}$ as $x$ approaches $4$ .
Substitute Value: Substitute the value of $x$ into the function to see if the function is defined at that point. $\lim_{x \to 4}\frac{3x}{(x-2)(x+2)} = \frac{3\cdot 4}{(4-2)(4+2)}$
Perform Calculations: Perform the calculations to evaluate the limit. $(3 \times 4)/((4-2)(4+2)) = 12/(2 \times 6) = 12/12 = 1$
Conclude Limit: Conclude the limit based on the calculations. $\newline$ Since we were able to directly substitute $x = 4$ into the function and get a result without any indeterminate forms, the limit exists and is equal to $1$ .