Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$ 
Calculate 
  $$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$

My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$

Use the fact  that
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$
to get that
\begin{align}
\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}
&=\lim_{x\to 3}\left(\frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}\times
\frac{(x+5)^{2/3}+2(x+5)^{1/3}+4}{(x+5)^{2/3}+2(x+5)^{1/3}+4}
\right)\\
&=\lim_{x\to 3}\left(\frac{\cos (\pi x)+1}{x-3}\times [(x+5)^{2/3}+2(x+5)^{1/3}+4]\right)\\
&=0\times 12=0.
\end{align}
since
$$
\lim_{x\to 3}\left(\frac{\cos (\pi x)-(-1)}{x-3}\right)
$$
is the definition of $f'(3)$ for $f(x)=\cos(\pi x)$. But
$$
f'(3)=-\pi\sin(3\pi)=0.
$$ 
