Limit of the Derivative

Since the integrand $\large \frac{x^4}{1+x^2}$ is always positive or $0$ , $f(x)$ is an increasing function and $f(x) > x$ for every $x$ . Also, $\large \frac{x^4}{1+x^2} \geq \frac{x^4}{2x^2}=\frac{x^2}{2}$ if $x \geq 1$ and since $\large \int_t^{t+1}\frac{x^2}{2}$ $dx > 2$ if $t$ is large (for example, if $t > 2$ ), we see that $f(x) < x + 1$ . Now by differentiating both sides of identity $\large \int_t^{f(t)}\frac{x^4}{1+x^2}$ $dx = 2$ , we see that $\large \frac{f(t)^4}{1+f(t)^2}$ $f'(t) = \large \frac{t^4}{1+t^2}$ where $t \leq f(t) \leq t+1$ . Therefore, $f'(t) = \large \frac{t^4}{f(t)^4} \cdot \frac{1+f(t)^2}{1+t^2}$ goes to 1 as $t$ goes to the infinity by the Sandwich Theorem (Squeeze Theorem).

The integrand function $\frac{x^4}{1+x^2}$ is positive, so that we must have $f(t) \ge t$ in order for the integral to give a positive result. Since the value of the integrand $\lim_{x \to_\infty}\frac{x^4}{1+x^2} = \infty$ , it must be that $\lim_{t \to_\infty} f(t)-t =0$ in order to keep the integral finite. From this it follows that $\lim_{t \to_\infty} f'(t) =1$ .