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 [tex]\frac{1}{6}[/tex].
Limit:
The limit given is:
[tex]\lim_{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7}[/tex]
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:
[tex]\sqrt{x+2} - 3[/tex]
It's conjugate is:
[tex]\sqrt{x+2}+3[/tex]
Multiplying numerator and denominator by the conjugate, meaning option A is correct:
[tex]\lim_{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x-7} \times \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}[/tex]
We do this because at the numerator we can apply:
[tex](a+b)(a-b) = a^2 - b^2[/tex]
Thus
[tex]\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)}[/tex]
Thus, we can simplify the factors of x - 7, meaning that option D is correct, and we get:
[tex]\lim_{x \rightarrow 7} \frac{1}{\sqrt{x+2}+3}[/tex]
Now, we just calculate the limit:
[tex]\lim_{x \rightarrow 7} \frac{1}{\sqrt{x+2}+3} = \frac{1}{\sqrt{7+2}+3} = \frac{1}{3+3} = \frac{1}{6}[/tex]
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 https://brainly.com/question/12207599
