<s>[INST] What is the following limit? $\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$ What is the following limit?
$$\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$$ Should I calculate the exact value of $(1+x)^5$?
 [/INST] It suffices to know the coefficient of $x^2$ in the numerator.
Note that $(1+x)^5-1-5x=\binom{5}{2}x^2+\mathcal{O}(x^3)$ and $x^2+x^5=x^2+\mathcal{O}(x^3)$ and hence the limit will be
$$\lim_{x \to 0} \frac{10x^2+\mathcal{O}(x^3)}{x^2+\mathcal{O}(x^3)}=10$$
Alternatively, if you don't want to use these symbols, just cancel out $x^2$ and obtain
$$\lim_{x \to 0} \frac{10+ax+bx^2+cx^3}{1+x^3}=10$$
Of course, it is not hard to compute the values of $a,b,c$ using binomial coefficients but it is not necessary for determining the limit.
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