<s>[INST] 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!
 [/INST] 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 $\infty$.
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