Question

Find the following limit. Limit of StartFraction StartRoot x + 2 EndRoot minus 3 Over x minus 7 EndFraction as x approaches 7 Which statements describe finding the limit shown? Check all that apply. Multiply by StartFraction StartRoot x + 2 EndRoot + 3 Over StartRoot x + 2 EndRoot + 3 EndFraction. Get x – 1 in the numerator. Get (x -7)(StartRoot x + 2 EndRoot minus 3) in the denominator. Divide out a common factor of x – 7. Calculate the limit as StartFraction 1 Over 6 EndFraction.
ANSWERS: A,D,E

Answer 1

Answer:

a,d,e

Step-by-step explanation:

Answer 2

In this question, we apply limit concepts to get the desired limit, finding that the correct options are: A, D and E, leading to the result of the limit being $\frac{1}{6}$.

Limit:

The limit given is:

$\lim_{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}$

If we apply the usual thing, of just replacing x by 7, the denominator will be 0, so this is not possible.

When we have a term with roots, we rationalize it, multiplying both the denominator and the denominator by the conjugate.

Multiplication by the conjugate:

The term with the root is:

$\sqrt{x+2} - 3$

It's conjugate is:

$\sqrt{x+2}+3$

Multiplying numerator and denominator by the conjugate, meaning option A is correct:

$\lim_{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} \times \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}$

We do this because at the numerator we can apply:

$(a+b)(a-b) = a^2 - b^2$

Thus

$\lim_{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} \times \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3} = \lim_{x \rightarrow 7} \frac{(\sqrt{x+2})^2 - 3^2}{(x-7)(\sqrt{x+2}+3)} = \lim_{x \rightarrow 7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)} = \lim_{x \rightarrow 7}\frac{x-7}{(x-7)(\sqrt{x+2}+3)}$

Thus, we can simplify the factors of x - 7, meaning that option D is correct, and we get:

$\lim_{x \rightarrow 7} \frac{1}{\sqrt{x+2}+3}$

Now, we just calculate the limit:

$\lim_{x \rightarrow 7} \frac{1}{\sqrt{x+2}+3} = \frac{1}{\sqrt{7+2}+3} = \frac{1}{3+3} = \frac{1}{6}$

Thus, option E is also correct.

Using a limit calculator, as given by the image below, we have that 1/6 is the correct answer.

For more on limits, you can check brainly.com/question/12207599