$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos (5x)}}{{\cos (7x) - 1}} = \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos (5x)} \right)\left( {1 + \cos (5x)} \right)\left( {\cos (7x) + 1)} \right)}}{{\left( {\cos (7x) - 1} \right)\left( {1 + \cos (5x)} \right)\left( {\cos (7x) + 1)} \right)}} = \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - {{\cos }^2}(5x)} \right)\left( {\cos (7x) + 1)} \right)}}{{\left( {{{\cos }^2}(7x) - 1} \right)\left( {1 + \cos (5x)} \right)}} = \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}(5x)\left( {\cos (7x) + 1)} \right)}}{{ - {{\sin }^2}(7x)\left( {1 + \cos (5x)} \right)}} = \cr
& \mathop {\lim }\limits_{x \to 0} - \frac{{\frac{{{{\sin }^2}(5x)}}{{{{(5x)}^2}}} \cdot 25{x^2} \cdot \left( {\cos (7x) + 1)} \right)}}{{\frac{{{{\sin }^2}(7x)}}{{{{(7x)}^2}}} \cdot 49{x^2} \cdot \left( {1 + \cos (5x)} \right)}} = \cr
& \mathop {\lim }\limits_{x \to 0} - \frac{{\left( {\frac{{\sin (5x)}}{{5x}}} \right) \cdot 25 \cdot \left( {\cos (7x) + 1)} \right)}}{{\left( {\frac{{\sin (7x)}}{{7x}}} \right) \cdot 49 \cdot \left( {1 + \cos (5x)} \right)}} = \cr
& \mathop {\lim }\limits_{x \to 0} - \frac{{1 \cdot 25 \cdot 2}}{{1 \cdot 49 \cdot 2}} = - \frac{{25}}{{49}} \cr} $
