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 $(1+x)^5$?
 A: 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.
A: Using $$\displaystyle (1+x)^5 = \binom{5}{0}+\binom{5}{1}x+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5$$
and Using $$\displaystyle \binom{n}{r} = \frac{n!}{r!\times (n-r)!}\;,$$ Where $n!=n\times (n-1)\times (n-2)\times ...2\times 1$
So we get $$\displaystyle (1+x)^5 = 1+5x+10x^2+10x^3+5x^4+x^5$$
So $$\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^5-1-5x}{x^2+x^5} = \lim_{x\rightarrow 0}\frac{10x^2+10x^3+5x^4+x^5}{x^2+x^5} = \lim_{x\rightarrow 0}\frac{10+10x+5x^2+x^3}{1+x^3} =10$$
A: HINT: we get $$(1+x)^5-1-5x=x^2(10+10x+5x^2+x^3)$$
