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!
 A: 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.
A: 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 $\infty$.
A: 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 
$$\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}\sim \frac{kx+\cdots}{x^2}\sim \frac k x$$
