Overpowered

$\large f(x)=\displaystyle \frac{\int_{x}^{x+24} \sqrt{t^2+1} dt}{x}$

Put $x=\infty$ in both neumerator and denominator..In neumerator evaluate integral by-parts put as limits infinity the value will be tends to infinity.Denominator will also be infinity.

Now differentiate w.r.t. x .For nuemerator use Newton-Liebnitz rule.

$f(x)=\displaystyle \lim_{x \to \infty}\frac{\sqrt{(x+24)^2+1} - \sqrt{x^2+1}}{1}$

Now substitute $x=\frac{1}{z}$

$\displaystyle \lim_{z \to 0} \frac{\sqrt{(1+24z)^2 + z^2}- \sqrt{z^2+1}}{z}$

putting $z=0$ in both sides it is in $0/0$ form .So using L'hospital Rule differentiating w.r.t. $x$

$\displaystyle \lim_{z \to 0} \frac{2(1+24 z)24 +2z}{2\sqrt{(1+24z)^2 + z^2}} - \frac{2 z}{2\sqrt{z^2+1}}$

putting $z=0$ we get the value $\boxed{24}$

$\begin{aligned} L & = \lim_{x \to \infty} \frac 1x \int_x^{x + 24} \sqrt{t^2 + 1} \ dt & \small \color{#3D99F6} \text{L'hopital's Rule} \\ & = \lim_{x \to \infty} \frac{d}{dx} \int_x^{x+24} \sqrt{t^2 + 1} \ dt \\ & = \lim_{x \to \infty} \frac{d}{dx} \left(\int_c^{x + 24} \sqrt{t^2 + 1} \ dt - \int_c^x \sqrt{t^2 + 1} \ dt \right) \\ & = \lim_{x \to \infty} \sqrt{(x + 24)^2 + 1} - \sqrt{x^2 + 1} & \small \color{#3D99F6} \text{Multiplying through by } \sqrt{(x + 24)^2 + 1} + \sqrt{x^2 + 1} \\ & = \lim_{x \to \infty} \frac{48x + 576}{\sqrt{(x + 24)^2 + 1} + \sqrt{x^2 + 1}} \\ & = \lim_{x \to \infty} \frac{48x\left(1 + \frac{12}{x}\right)}{x\sqrt{1 + \frac{48}{x} + \frac{577}{x^2}} + x\sqrt{1 + \frac{1}{x^2}}} \\ & = \lim_{x \to \infty} \frac{48x}{2x} \\ & = \boxed{24} \end{aligned}$