Find the limit of 9 minus x squared as x approaches positive infinity.

Step-by-step solution

1. 1

   Rewrite so the leading term is visible. Reordering, $9-x^2=-x^2+9$: this is a quadratic with leading coefficient $-1$. The degree-two term always wins the race at infinity, so its sign decides the answer.

2. 2

   Evaluate the leading term. $x^2\to+\infty$ as $x\to\infty$, therefore

   $-x^2\to-\infty$

   The minus sign in front is the crux — students often report $+\infty$ by looking only at the $x^2$.

3. 3

   Add the constant. Adding a fixed $9$ to a quantity falling below every bound changes nothing about the divergence:

   $9-x^2\to-\infty$

4. 4

   Confirm numerically. At $x=10$: $9-100=-91$. At $x=100$: $-9991$. At $x=1000$: $-999991$. The outputs decrease without limit.

5. 5

   Connect to the graph. $y=9-x^2$ is a parabola opening downward with vertex $(0,9)$, so both arms fall forever; indeed $\lim_{x\to-\infty}(9-x^2)=-\infty$ as well. The limit is $-\infty$, meaning the function diverges rather than settling on any finite value.