Limit exercise solved without using series or l'hopital rule , idea of solution credit to Maher Albaron

Exercise:

Show that, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-x-1}{{{x}^{2}}}=\frac{1}{2}$ without using the hospital rule .

Solution: Let $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-x-1}{{{x}^{2}}}$   as $x\to 0\,\,,\,\,\,2x\to 0$

Let \(L=\underset{2x\to 0}{\mathop{\lim }}\,\frac{{{e}^{2x}}-2x-1}{{{\left( 2x \right)}^{2}}}=\underset{2x\to 0}{\mathop{\lim }}\,\frac{{{\left( {{e}^{x}} \right)}^{2}}\overbrace{-2{{e}^{x}}+2{{e}^{x}}}^{0}\overbrace{-2+1}^{-1}-2x}{{{\left( 2x \right)}^{2}}}\)

$=\underset{2x\to 0}{\mathop{\lim }}\,\frac{{{\left( {{e}^{x}}-1 \right)}^{2}}+2{{e}^{x}}-2-2x}{4{{x}^{2}}}=\underset{2x\to 0}{\mathop{\lim }}\,\frac{{{\left( {{e}^{x}}-1 \right)}^{2}}+2\left( {{e}^{x}}-1-x \right)}{4{{x}^{2}}}$

$\Rightarrow L=\underset{2x\to 0}{\mathop{\lim }}\,{{\left( \frac{{{e}^{x}}-1}{2x} \right)}^{2}}+\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-x-1}{{{x}^{2}}}=\underset{2x\to 0}{\mathop{\lim }}\,{{\left( \frac{{{e}^{x}}-1}{2x} \right)}^{2}}+\frac{1}{2}L$

$\Rightarrow \frac{1}{2}L=\frac{1}{4}\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{{{e}^{x}}-1}{x} \right)}^{2}}\Leftrightarrow L=\frac{1}{2}$  (since $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{0}}}{x-0}=\frac{d}{dx}{{\left( {{e}^{x}} \right)}_{x=0}}={{e}^{0}}=1$ )