Light Newton Binomial

$\begin{aligned} L & = \lim_{x \to 0} \frac {{\color{#3D99F6}(1+x)^5}-(1+5x)}{x^2+x^5} & \small \color{#3D99F6} \text{Expand }(1+x)^5 \\ & = \lim_{x \to 0} \frac {{\color{#3D99F6}1+5x + 10x^2 + 10x^3 + 5x^4 + x^5}-(1+5x)}{x^2+x^5} \\ & = \lim_{x \to 0} \frac {10x^2 + 10x^3 + 5x^4 + x^5}{x^2+x^5} & \small \color{#3D99F6} \text{Divide up and down by }x^2. \\ & = \lim_{x \to 0} \frac {10 + 10x + 5x^2 + x^3}{1+x^3} \\ & = \boxed{10} \end{aligned}$

$\begin{aligned} L &= \lim_{x\to\infty}\frac{(1+x)^5-(1+5x)}{x^2+x^5} &\text{Note that if we plug in 0, we get 0/0. This means we can use L'Hôpital's rule.} \\ &= \lim_{x\to\infty}\frac{5(1+x)^4-5}{2x+5x^4} \\ &= \lim_{x\to\infty}\frac{20(1+x)^3}{2+20x^3} \\ &= \lim_{x\to\infty}\frac{20}{2} \\ &= \boxed{10} \end{aligned}$

$L=\lim_{x \to 0} \frac{(1+x)^{5} - (1+5x)}{x^{2}+x^{5}}$ $L=\lim_{x \to 0} \frac{1+5x+C_{5}^{2} x^{2}+o(x^{2}) - (1+5x)}{x^{2}+x^{5}}$ $L=\lim_{x \to 0} \frac{ 5\times 4\div 2x^{2}+o(x^{2}) }{x^{2}+x^{5}}$ $L=10$