What is \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}}\) equal to

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}}\)                 [Form 0/0]

Apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {0 + 1} \right)}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;1}}{{2x}}\)              [Form 0/0]

Again apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;0}}{2} = \frac{{{e^0}}}{2} = \frac{1}{2}\)