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guajiro [1.7K]
2 years ago
13

What is limit of startfraction startroot x + 2 endroot minus 3 over x minus 7 endfraction as x approaches 7? negative one-sixth

0 one-sixth dne
SAT
2 answers:
Ne4ueva [31]2 years ago
4 0

Answer:

c 1/6

Explanation:

Vladimir [108]2 years ago
3 0

The value of  limit of startfraction startroot x + 2 endroot minus 3 over x minus 7 endfraction when x approaches 7 is 1/6.

<h3>How to evaluate a function for a given value?</h3>

To evaluate a function for a particular value, plug in the value of the variable in the function.

The expression given in the problem is,

\lim_{x \to 7}  \dfrac{\sqrt{(x+2)}-3}{x-7}

Multiply the factor √(x+2)+3 above and below of fraction,

\lim_{x \to 7}  \dfrac{\sqrt{(x+2)}-3}{x-7}\times\dfrac{\sqrt{(x+2)}+3}{\sqrt{(x+2)}+3}\\\lim_{x \to 7}  \dfrac{\sqrt{(x+2)}^2-3^2}{(x-7)(\sqrt{(x+2)}+3)}\\\lim_{x \to 7} \dfrac{x+2-9}{(x-7)(\sqrt{(x+2)}+3)}\\\lim_{x \to 7} \dfrac{x-7}{(x-7)(\sqrt{(x+2)}+3)}\\\lim_{x \to 7} \dfrac{1}{\sqrt{(x+2)}+3)}

The value of x tents to 7. Thus,

}\dfrac{1}{\sqrt{7+2)}+3}\\

1/(3+3)=1/6

Thus, the value of  limit of startfraction startroot x + 2 endroot minus 3 over x minus 7 endfraction when x approaches 7 is 1/6.

Learn more about the evaluate a function here;

brainly.com/question/2284360

#SPJ1

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