Evaluate $\lim_{x\to 0} \frac{(1-x)^{1/3}-(1+x)^{1/2}}{x}$ Evaluate the limit
$$\lim_{x\to 0} \frac{(1-x)^{1/3}-(1+x)^{1/2}}{x}$$
I know the limit is $-5\over6$ by looking at the graph of the function, but how can I algebraically show that that is the limit?
 A: One method here is to use L'Hopital's rule.  We know that $(1-x)^{1/3}-(1+x)^{1/2}$ and $x$ tend to $0$ as $x\to0$; so,
$$
\lim_{x\to 0}\frac{(1-x)^{1/3}-(1+x)^{1/2}}{x}=\lim_{x\to0}\frac{\frac{d}{dx}\left[(1-x)^{1/3}-(1+x)^{1/2}\right]}{\frac{d}{dx}[x]}
$$
(because this last limit exists, of course!).
Alternatively, you can solve this using the definition of the derivative.  Let $f(x)=(1-x)^{1/3}$ and $g(x)=(1+x)^{1/2}$. Then adding/subtracting $1$ in the numerator yields
$$
\lim_{x\rightarrow0}\frac{(1-x)^{1/3}-(1+x)^{1/2}}{x}=\lim_{x\rightarrow0}\left[\frac{(1-x)^{1/3}-1}{x-0}-\frac{(1+x)^{1/2}-1}{x-0}\right].
$$
But, if you look at it, we've just written
$$
\lim_{x\rightarrow0}\left[\frac{f(x)-f(0)}{x-0}-\frac{g(x)-g(0)}{x-0}\right]=f'(0)-g'(0).
$$
A: or simply 
$$
\lim_{x\to 0}\frac{(1-x)^{1/3}+1-1-(1+x)^{1/2}}{x}=\lim_{x\to0}-\frac{(1-x)^{1/3}-1}{-x}-\frac{(1+x)^{1/2}-1}{x}=-\frac{5}{6}
$$
A: Using the binomial expansions,
$$(1-x)^{1/3}=1-\frac{1}{3}x+O(x^{2}) \mbox{  and  } (1+x)^{1/2}=1+\frac{1}{2}x+O(x^{2})$$
So your expression is simply $$\lim_{x \to 0}\frac{\frac{-5}{6}x+O(x^{2})}{x} = -5/6$$
A: Use L'Hopital's rule (both numerator and denominator go to zero).
