Question

what value does $f(x) = \frac{1}{(x - 7)^2}$ approach as $x \to 7$? (use $\infty$ or $-\infty$ if appropriate. enter dne if the limit is not infinite and does not exist.) answer: submit answer next item

Explanation:

Step1: Analyze the function's denominator

The function is \( f(x)=\frac{1}{(x - 7)^2} \). As \( x
ightarrow7 \), we look at the behavior of the denominator \( (x - 7)^2 \). When \( x \) gets very close to 7, \( x-7 \) gets very close to 0, and squaring a number close to 0 (whether positive or negative, since we square it) makes the denominator a positive number that approaches 0.

Step2: Analyze the behavior of the whole function

Since the numerator is 1 (a constant) and the denominator approaches 0 from the positive side (because it's a square, so \( (x - 7)^2>0 \) for \( x
eq7 \)), the value of \( \frac{1}{(x - 7)^2} \) will become very large (tend to positive infinity) as \( x \) approaches 7. Because when you divide a constant by a number that gets closer and closer to 0 (while staying positive), the result grows without bound.

Answer:

\(\infty\)