Question: Find the limit of $$\begin{equation\*}
\lim\_{x \rightarrow 0}
\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}
\end{equation\*}$$

Shall I use the binomial theorem? 
Any hint will be appreciated!

Answer: 
Ok, now that OP's conditions are better known, let's try again.

Numerator is 
$$(x-1)^5+(1+5x)\equiv x^5-5x^4+10x^3-10x^2+5x-1+(1+5x)$$$$\equiv x^5-5x^4+10x^3-10x^2+10x$$

so the given expression is

$$\frac{x^5-5x^4+10x^3-10x^2+10x}{x^5+x^2}\equiv\frac{x^4-5x^3+10x^2-10x+10}{x^4+x}.$$

Now, numerator does not approach $0$ as $x\to 0$ but denominator does, so the limit does not exist.
Considering $$\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}$$ the $1$ will cancel out (because of power $5$ in numerator) and you are left with terms in $x$ for the lowest degree in numerator. So 
$$\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}\sim \frac{kx+\cdots}{x^2}\sim \frac k x$$