Which of the following function's output decreases without bound as $x$ increases without bound?
$f(x)=x^{3}-117$
$f(x)=1-x^{2}+x^{3}$
$f(x)=x^{7}-x^{5}-x^{3}-10$
$f(x)=x^{2}-10 x-24$
$f(x)=1-x^{2}$

Step 1 ：We need to find the function whose output decreases without bound as $x$ increases without bound. This means that as $x$ becomes larger and larger, the value of the function becomes smaller and smaller (negative) and continues to do so indefinitely.

Step 2 ：To find this, we need to look at the highest degree term in each function. This is because, for large values of $x$, the highest degree term will dominate the value of the function.

Step 3 ：For the first function, $f(x)=x^{3}-117$, the highest degree term is $x^{3}$, which increases without bound as $x$ increases.

Step 4 ：For the second function, $f(x)=1-x^{2}+x^{3}$, the highest degree term is $x^{3}$, which also increases without bound as $x$ increases.

Step 5 ：For the third function, $f(x)=x^{7}-x^{5}-x^{3}-10$, the highest degree term is $x^{7}$, which increases without bound as $x$ increases.

Step 6 ：For the fourth function, $f(x)=x^{2}-10 x-24$, the highest degree term is $x^{2}$, which increases without bound as $x$ increases.

Step 7 ：For the fifth function, $f(x)=1-x^{2}$, the highest degree term is $-x^{2}$, which decreases without bound as $x$ increases.

Step 8 ：So, the function whose output decreases without bound as $x$ increases without bound is $f(x)=1-x^{2}$.

Step 9 ：\(\boxed{f(x)=1-x^{2}}\) is the function whose output decreases without bound as $x$ increases without bound.