How to calculate limit? I'm puzzled with this limit. The answer is -0.5, but how to get it? $\lim_\limits{x \to \infty}1-x+\sqrt{\frac{x^3}{x+3}}$
 A: A start: The interesting part is $\sqrt{\frac{x^3}{x+3}}-x$. Consider
$$\frac{\left(\sqrt{\frac{x^3}{x+3}}-x\right)  \left(\sqrt{\frac{x^3}{x+3}}+x\right) }{ \sqrt{\frac{x^3}{x+3}}+x }.$$
A: Here is a totally different approach: Set $\sqrt{\frac{x}{x+3}}=t$ and solve for $x$ to find $x=\frac{3t^2}{1-t^2}$. Put this back into your limit to find $1-\frac{3t^2}{1-t^2} + \frac{3t^3}{1-t^2}$ where $t$ goes to $1$. This is indeterminate and so put everything into one fraction: $\frac{1-4t^2+3t^3}{1-t^2}$ which for $t→1$ now becomes "zero over zero". Apply L'Hospital's Rule to find $-0.5$
A: I've tried to multiply by conjugate not only part with a variable, but all the expression, and this way I calculate it without replace and L'Hospital's Rule.
$\lim_\limits{x \to \infty} 1-x+\sqrt{\frac{x^3}{x+3}}=\lim_\limits{x \to \infty}\frac{\frac{x^3}{x+3}-(x-1)^2}{\sqrt{\frac{x^3}{x+3}}+x-1} = \lim_\limits{x \to \infty}\frac{-x^2+5x-3}{(x+3)\left(\frac{\sqrt{x^3}+\sqrt{x+3}(x-1)}{\sqrt{x+3}}\right)}=\lim_\limits{x \to \infty}\frac{-x^2+5x-3}{\sqrt{x^4+3x^3}+(x+3)(x+1)}$
Now we can see easily that factor before $x^2$ is $-1$ in numerator and $2$ in denominator, so the result is $-\frac{1}{2}$
