<s>[INST] 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 :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
 [/INST] 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
$$\lim_{t\to 0} \frac {-\cos (\pi t)+1}{2\left(\left(1+\frac{t}{8}\right)^{1/3} -1\right)}=\lim_{t\to 0} \frac {\frac{(\pi t)^2}{2} +o(t^2)}{2\left(\frac{t}{24}+o(t)\right)}=\lim_{t\to 0} \frac {\frac{\pi^2 t}{2} +o(t)}{\frac{1}{12}+o(1)}=0.$$
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