Show that $\lim\limits_{x\to\infty} f(x)$ exists if $f'(x) = \frac{1}{x^2+f(x)^2}$ and $f(1)=1$ 
Let $f(x)$ be a real differentiable function defined for $x\geq 1$ such that $f(1)=1$ and $f'(x)=\dfrac{1}{x^2+f(x)^2}.$ Show that $$\lim_{x\to \infty}f(x)$$ exists and is less than $1+\frac{\pi}{4}$

I have no idea how to tackle this question... Help!
 A: Notice that $f'(x)$ is always positive, so $f$ is increasing. Hence for $x > 1$ we have $f(x) > 1$. Consequently, for $x > 1$, we have $f'(x) < \frac{1}{x^2 + 1}$.
We also have that $f'(x)$ is continuous for $x > 1$ since it's the quotient of continuous functions and the denominator is not zero. (The numerator is the continuous function g(x) = 1, and the decnominator is the sum of continuous functions $h(x) = x^2$ and $f(x)$.) Thus, we can apply the fundamental theorem of calculus, which tells us
$$
  \int_1^t f'(x)dx = f(t) - f(1).
$$
for any $t > 1$.
For any $t > 1$, we then have
$$
\begin{align*}
  f(t) 
  &= f(1) + f(t) - f(1) \\
  &= f(1) + \int_1^t f'(x)dx \\
  &< f(1) + \int_1^t \frac{1}{x^2 + 1}dx \\
  &= f(1) + \left.\tan^{-1}(x)\right\rvert_{x=1}^{x=t} \\
  &< 1 + \frac{\pi}{4}.
\end{align*}
$$
Since $f(t)$ is increasing and is bounded above, the limit exists.
Then, taking the limit as $t$ approaches infinity, we have
\begin{align*}
  \lim_{t \to \infty} f(t) 
  &= f(1) + \lim_{t \to \infty} \int_1^t f'(x)dx \\
  &< f(1) + \lim_{t \to \infty} \int_1^t \frac{1}{x^2 + 1}dx \\
  &= 1 + \frac{\pi}{4}
\end{align*}
as desired.
