The value of limit of startfraction startroot x + 2 endroot minus 3 over x minus 7 endfraction when x approaches 7 is 1/6.
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,
[tex]\lim_{x \to 7} \dfrac{\sqrt{(x+2)}-3}{x-7}[/tex]
Multiply the factor √(x+2)+3 above and below of fraction,
[tex]\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)}[/tex]
The value of x tents to 7. Thus,
[tex]}\dfrac{1}{\sqrt{7+2)}+3}\\[/tex]
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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