Question

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

Answer 1

Answer:

c 1/6

Explanation:

Answer 2

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

How to evaluate a function for a given value?

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.

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