\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1}}{{\sin x - x}}\) is 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.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

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

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Given:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x\; -\;1}}{{sin\;x\; - \;x}} = \left( {\frac{0}{0}} \right)\) form

Applying L’ Hospital  rule:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x - 1}}{{sin\;x-\;x}} =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({cos\;x\; -\;1})}{{\frac{d}{dx}(sin\;x\;-\;x})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\;x {\rm{}}}}{{\cos {\rm{x}\;-\;1}}} \)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\; {\rm{x}}}}{{\cos {\rm{x}\,-1\;}}} = \left( {\frac{0}{0}} \right){\rm{form}}\)

Once again by L’ Hospital rule,

\({\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-cos\; {\rm{x}}}}{{{\rm{-sin~x}}}} = \frac{-1}{0} = \infty{\rm{}}\)