Question

lim x tends to 0: 1-cos2mx÷ 1- cos2nx

Answer 1

here is your answer. thank you

Answer 2

The solution of the given limit is

      $\lim_{x \to 0\1} \frac{1-cos2mx}{1-cos2nx} = \frac{m^{2} }{n^{2} }$

• here  We have,  

               $\lim_{x \to 0\1} \frac{1-cos2mx}{1-cos2nx}$    

• Now, by using double angle formula for cosx that is

            $cos2x = 1-2sin^{2}x\\ 1 - cos2x = 2sin^{2} x$

   

           we get,  

                    $\lim_{x \to 0\1} \frac{2sin^{2}mx }{2sin^{2}nx }$             

•  dividing both numerator and denominator by '$x^{2}$' and multiplying and dividing both numerator and denominator by $m^{2}$ and $n^{2}$ respectively, we get

                 $=\lim_{x \to 0\1} \frac {\frac{m^{2} sin^{2} mx}^{m^{2} } {x^{2} } } { {\frac{n^{2} sin^{2} nx}^{n^{2} } {x^{2} } } }$

                 $=\frac{m^{2} }{n^{2} } \frac{ \lim_{x \to 0\1} (\frac{sinmx}{mx} )^{2} } { \lim_{x \to 0\1}(\frac{sinnx}{nx} )^{2} }$

• Now, by using standard sine limit that is

               $\lim_{x \to 0\1} \frac{sinx}{x} = 1$    , we get

                $\frac{m^{2} }{n^{2} } \frac{ \lim_{x \to 0\1} (\frac{sinmx}{mx} )^{2} }{ \lim_{x \to 0\1} (\frac{sinnx}{nx} )^{2}} = \frac{m^{2} }{n^{2} } (\frac{1}{1} )$

       

Therefore the required solution of the given limit is ,

        $\lim_{x \to 0\1} \frac{1-cos2mx}{1-cos2nx} = \frac{m^{2} }{n^{2} }$