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Question: \[\lim_{x \rightarrow 0}\frac{1 - \cos mx}{1 - \cos nx} =\]...

lim⁡x→01−cos⁡mx1−cos⁡nx=\lim_{x \rightarrow 0}\frac{1 - \cos mx}{1 - \cos nx} =

A

m/nm/n

B

n/mn/m

C

m2n2\frac{m^{2}}{n^{2}}

D

n2m2\frac{n^{2}}{m^{2}}

Answer

m2n2\frac{m^{2}}{n^{2}}

Explanation

Solution

lim⁡x→01−cos⁡mx1−cos⁡nx=lim⁡x→0{2sin⁡2mx22sin⁡2nx2}\lim_{x \rightarrow 0}\frac{1 - \cos mx}{1 - \cos nx} = \lim_{x \rightarrow 0}\left\{ \frac{2\sin^{2}\frac{mx}{2}}{2\sin^{2}\frac{nx}{2}} \right\}

=lim⁡x→0[{sin⁡mx2mx2}2m2x24.1{sin⁡nx2nx2}2.4n2x2]=m2n2×1=m2n2= \lim_{x \rightarrow 0}\left\lbrack \left\{ \frac{\sin\frac{mx}{2}}{\frac{mx}{2}} \right\}^{2}\frac{m^{2}x^{2}}{4}.\frac{1}{\left\{ \frac{\sin\frac{nx}{2}}{\frac{nx}{2}} \right\}^{2}}.\frac{4}{n^{2}x^{2}} \right\rbrack = \frac{m^{2}}{n^{2}} \times 1 = \frac{m^{2}}{n^{2}}

Trick : Apply L-Hospital rule ,

lim⁡x→01−cos⁡mx1−cos⁡nx=lim⁡x→0msin⁡mxnsin⁡nx\lim_{x \rightarrow 0}\frac{1 - \cos mx}{1 - \cos nx} = \lim_{x \rightarrow 0}\frac{m\sin mx}{n\sin nx} =lim⁡x→0m2cos⁡mxn2cos⁡nx=m2n2.= \lim_{x \rightarrow 0}\frac{m^{2}\cos mx}{n^{2}\cos nx} = \frac{m^{2}}{n^{2}}.