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.
Divide top and bottom by $x^2$ to get $$\lim\_{x \to 0} \frac{\frac{1}{x^2}(1+5x) + \frac{1}{x^2}(x-1)^5}{1+x^3}$$

This is a quotient where the denominator has limit $1$, so the limit exists iff the following limit exists, and it has the same value as this limit: $$\lim\_{x \to 0} \left[\frac{1}{x^2}(1+5x) + \frac{1}{x^2}(x-1)^5\right]$$

Now this can be done using the binomial theorem; alternatively, substitute $u=x^2$ to get $$\lim\_{u \to 0} \left[\frac{(1+5\sqrt{u})+(\sqrt{u}-1)^5}{u} \right]$$
which is the derivative of the function $y \mapsto (1+5 \sqrt{y}) + (\sqrt{y} - 1)^5$ at $y=0$; that derivative is easily evaluated and its limit to $0$ taken, and that limit is manifestly $\infty$.