Respuesta :

Answer:

Hence, the limit of the expression [tex]\lim_{x \to 3} \dfrac{x-3}{x^2-9}[/tex] is:

0.1667 (i.e. option b is true)

Step-by-step explanation:

We are asked to estimate the limit of the expression:

[tex]\lim_{x \to 3} \dfrac{x-3}{x^2-9}[/tex]

We know that:

[tex]a^2-b^2=(a-b)(a+b)[/tex]

Hence, we could represent it as:

[tex]\lim_{x \to 3} \dfrac{x-3}{x^2-3^2}\\ \\= \lim_{x \to 3} \dfrac{x-3}{(x-3)(x+3)}\\ \\= \lim_{x \to 3} \dfrac{1}{x+3}[/tex]

Since we cancel out the similar terms in the numerator as well as in the denominator.

[tex]\lim_{x \to 3} \dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}=0.1667[/tex]

Hence, the limit of the expression [tex]\lim_{x \to 3} \dfrac{x-3}{x^2-9}[/tex] is:

0.1667

zainebamir540

Answer:

Choice b is the answer.

Step-by-step explanation:

We have given a function.

f(x) = x-3 / x²-9

We have to find the limit of given function at x.

[tex]\lim_{x \to \ 3} x-3/x^{2} -9[/tex]

Applying difference formula to denominator of given function.

x-3 / x²-9 = x-3 / (x-3)(x+3)

x-3 / x²-9 =  1 / x+3

Applying limit, we have

[tex]\lim_{x \to \ 3} x-3/x^{2} -9[/tex]  =  [tex]\lim_{x \to \ 3} 1/x+3[/tex]

[tex]\lim_{x \to \ 3} x-3/x^{2} -9[/tex]  = 1/3+3

[tex]\lim_{x \to \ 3} x-3/x^{2} -9[/tex]  = 1/6

[tex]\lim_{x \to \ 3} x-3/x^{2} -9[/tex]  = 0.1667 which is the answer.

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