Calculating $\lim_{x\to 0}\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt{\log(x+1)}}\right)$ Find the limit 
$$\lim_{x\to 0}\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt{\log(x+1)}}\right)$$
 A: Hint 
$$\frac{\log(1+x)}{x}\to 1$$
$$\frac{\log(1+x)-x}{x^2}\to -\frac{1}{2}$$
Further Hint
$$\displaylines{
  \frac{1}{{\sqrt x }} - \frac{1}{{\sqrt {\log (x + 1)} }} = \frac{{\sqrt {\log (x + 1)}  - \sqrt x }}{{\sqrt {x\log \left( {1 + x} \right)} }}\left( {\frac{{\sqrt {\log (x + 1)}  + \sqrt x }}{{\sqrt {\log (x + 1)}  + \sqrt x }}} \right) \cr 
   = \frac{{\log (x + 1) - x}}{{\sqrt {x\log \left( {1 + x} \right)} }}{\left( {\sqrt {\log (x + 1)}  + \sqrt x } \right)^{ - 1}} \cr 
   = \frac{{\log (x + 1) - x}}{{\sqrt {x\log \left( {1 + x} \right)} }}\frac{1}{{\sqrt x }}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}}  + 1} \right)^{ - 1}} \cr 
   = \frac{1}{{\sqrt x }}\frac{{\log (x + 1) - x}}{x}{\left( {\sqrt {\frac{{\log \left( {1 + x} \right)}}{x}} } \right)^{ - 1}}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}}  + 1} \right)^{ - 1}} \cr 
   = \sqrt x \frac{{\log (x + 1) - x}}{{{x^2}}}{\left( {\sqrt {\frac{{\log \left( {1 + x} \right)}}{x}} } \right)^{ - 1}}{\left( {\sqrt {\frac{{\log (x + 1)}}{x}}  + 1} \right)^{ - 1}} \cr} $$
A: One more way: after a bit of algebra you get 
$$
\lim_{x \to 0}\frac{\log (x+1)-x}{\sqrt{x \log (x+1)}(\sqrt{\log (x +1)}+\sqrt{x})}
$$
Now expand in Maclaurin series: $\log (x+1) =x +O(x^2)$ to get 
$$
\frac{O(x)}{(1+O(x))\sqrt{(1+O(x)}}=0
$$
