find the limit of the following function without using L'Hopital's Rule

Question

Can help to solve the following limit without using L'Hopital's Rule?

$$\lim_{x\to 0}\frac{e^x-x-1}{cosx-1}$$

Answer 1

Use taylor series$$\lim_{x\to 0}\frac{e^x-x-1}{cosx-1}=\\\lim_{x\to 0}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-x-1}{(1-\frac{x^2}{2}+\frac{x^4}{4!}-...)-1}=\\$$can you go on ? $$\lim_{x\to 0}\frac{\frac{x^2}{2!}+\frac{x^3}{3!}+...}{(-\frac{x^2}{2}+\frac{x^4}{4!}-...)}=\\ \lim_{x\to 0}\frac{x^2(\frac{1}{2!}+\frac{x^1}{3!}+...)}{x^2(-\frac{1}{2}+\frac{x^2}{4!}-...)}=\\$$simplify $x^2$ $$\lim_{x\to 0}\frac{(\frac{1}{2!}+\frac{x^1}{3!}+...)}{(-\frac{1}{2}+\frac{x^2}{4!}-...)}=\\\frac{\frac12}{-\frac12}=-1$$

Answer 2

$$e^x=1+x+\frac{x^2}{2!}+o(x^2)\implies e^x-1-x=\frac{x^2}{2!}+o(x^2)$$

$$\cos(x)=1-\frac{x^2}{2!}+o(x^2)\implies \cos(x)-1=-\frac{x^2}{2!}$$

Therefore, $$\frac{e^x-1-x}{\cos(x)-1}=\frac{1+\varepsilon_1(x)}{-1+\varepsilon_2(x)}$$ where $\varepsilon_i$ are function s.t. $$\lim_{x\to 0}\varepsilon_i(x)=0.$$