The superlimit!

There are numerous ways to solve a limit based problem. My favorite has always been expansion.You always have the power to change any function into algebric function as $f(x) = f(0) + \frac{f'(0)}x + \frac{f''(0}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots$

In this solution, the dots will mean higher powers of $x$ . We know that $\sqrt{1+x^2} = 1 + \frac{x^2}{2} + \dots$

Hence, $\sqrt{1+x^2 } - x = 1 + ( - x + \frac{x^2}{2} + \dots)$

Also $\ln (1+ z) = z -\frac{z^2}{2} + \frac{z^3}{3} \dots$

Thus our expression becomes :

$\displaystyle \lim_{x \to 0} \bigg(\frac{1}{x - \frac{x^2}{2} + \dots} + \frac{1}{ (- x + \frac{x^2}{2} + \dots) - \frac{(- x + \frac{x^2}{2} + \dots)^2}{2} \dots} \bigg)$

$\displaystyle \lim_{x \to 0} \bigg(\frac{1}{x - \frac{x^2}{2} + \dots} + \frac{1}{-x + 0x^2 + \dots}$

= $\displaystyle \lim_{x \to 0} \frac{x - \frac{x^2}{2} - x + 0x^2 + \dots}{-x^2 + \dots}$

= $\boxed{\frac{1}{2}}$ . Hence, our answer is $50$