Provide an appropriate response.
-Let $f ( n ) = \sqrt { n ^ { 2 } + 9 } - \sqrt { n ^ { 2 } - 9 }$. What is $f ( n )$ approximately equal to as $n$ gets large? Hint: Compute various examples on your calculator.
A) $\frac { 9 } { n ^ { 2 } }$
B) $\frac { 9 } { 2 n }$
C) $\frac { 9 } { \sqrt { n } }$
D) $\frac { 9 } { \mathrm { n } }$

Question

Quizplus · Accepted Answer

After calculating various examples on a calculator, it becomes clear that as n gets large, the term involving the square root of $n^2$ dominates both the numerator and denominator of $f(n)$. Thus, we can ignore the term involving the constant '9' and simplify $f(n)$ to $\frac{9}{n}$, making choice D the best choice.

Provide an Appropriate Response f(n)=n2+9−n2−9f ( n ) = \sqrt { n ^ { 2 } + 9 } - \sqrt { n ^ { 2 } - 9 }f(n)=n2+9​−n2−9​

Provide an Appropriate Response $f ( n ) = \sqrt { n ^ { 2 } + 9 } - \sqrt { n ^ { 2 } - 9 }$