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 :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
 A: Your approach is not correct. You are using the expansions at $0$ of the functions $\cos(\pi x)$ and the $(x+5)^{1/3}$, whereas the limit is for $x$ that goes to $3$.
Note that by letting $x=t+3$, we have that $t\to 0$ and
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}=\lim_{t\to 0} \frac {\cos (\pi t +3\pi)+1}{(t+8)^{1/3} -2}=\lim_{t\to 0} \frac {-\cos (\pi t)+1}{2\left(\left(1+\frac{t}{8}\right)^{1/3} -1\right)}.$$
Now we can use the "known" expansions at $0$:
$$\cos(\pi t)=1-\frac{(\pi t)^2}{2} +o(t^2)\quad \text{and}\quad
\left(1+\frac{t}{8}\right)^{1/3}=1+\frac{t}{24}+o(t).$$
Hence, the limit becomes
$$\lim_{t\to 0} \frac {-\cos (\pi t)+1}{2\left(\left(1+\frac{t}{8}\right)^{1/3} -1\right)}=\lim_{t\to 0} \frac {\frac{(\pi t)^2}{2} +o(t^2)}{2\left(\frac{t}{24}+o(t)\right)}=\lim_{t\to 0} \frac {\frac{\pi^2 t}{2} +o(t)}{\frac{1}{12}+o(1)}=0.$$
A: 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
$$
f'(3)=-\pi\sin(3\pi)=0.
$$ 
A: $$\lim_{x\to 3} \dfrac {\cos (\pi x)+1}{(x+5)^{1/3} -2}=\frac{\pi^2}{2}\lim_{x\rightarrow3}\left(\frac{\sin^2\frac{\pi(x-3)}{2}}{\frac{\pi^2(x-3)^2}{4}}\cdot\frac{(x-3)^2}{\sqrt[3]{x+5}-2}\right)=$$
$$=\frac{\pi^2}{2}\lim_{x\rightarrow3}(x-3)\left(\sqrt[3]{(x+5)^2}+2\sqrt[3]{x+5}+4\right)=0.$$
A: With substitution $x=u+3$ we have
\begin{align}
\lim_{x\to 3} \dfrac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
&= \lim_{u\to 0}\dfrac{1-\cos\pi u}{\sqrt[3]{u+8}-2} \\
&= \lim_{u\to 0}\dfrac{2\sin^2\frac12\pi u}{u^2}\dfrac{u^2}{\sqrt[3]{u+8}-2}\dfrac{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4}{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4} \\
&= \dfrac{\pi^2}{2}\lim_{u\to 0}\dfrac{u^2}{u}\lim_{u\to 0}\dfrac{\sqrt[3]{u+8}^2+\sqrt[3]{u+8}+4}{1} \\
&= \dfrac{\pi^2}{2}\times 0 \times 12 \\
&= 0
\end{align}
A: By definition of derivative of a  function at $x=3$ we have, 
$$\text{since $\cos3\pi =-1\implies$ }\lim_{x\to 3} \frac {\cos (\pi x)-(-1)}{x-3} =\lim_{x\to 3} -\pi\sin (\pi x) =0$$
and 
$$\lim_{x\to 3} \frac {(x+5)^{\frac {1}{3}} -2}{x-3} =\lim_{x\to 3} \frac {1}{3}(x+5)^{\frac {-2}{3}} =\frac {1}{3} 8^{-2/3} =\frac {1}{12}.$$
Therefore, $$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}=\lim_{x\to 3} \frac {\cos (\pi x)+1}{\color{red}{x-3}}\frac {\color{red}{x-3}}{(x+5)^{\frac {1}{3}} -2}=0$$
