Determining the limit of a function containing $(x-1)^5$ without l'hopital`s rule. 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!

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.
