Question

$\displaystyle \lim_{x \to 0}$ $\frac{1-cos\,mx}{1-cos\,nx}=$

• A. $m/n$
• B. $m^2/n^2$
• C. $0$
• D. $n^2/m^2$

Answer 1

Answer: (B)

$m^2/n^2$

Answer 2

$\displaystyle \lim_{x \to 0}$ $\frac{1-cos\,mx}{1-cos\,nx}$ $=\displaystyle \lim_{x \to 0}$ $\frac{2\,sin^{2}\left(mx/2\right)}{2\,sin^{2}\left(nx/2\right)}$ $=\frac{\displaystyle \lim_{x \to 0}\left(\frac{sin\left(mx/2\right)}{\left(mx/2\right)}\right)^{2}\cdot\frac{m^{2}x^{2}}{4} }{\displaystyle \lim_{x \to 0} \left(\frac{sin\left(nx/2\right)}{\left(nx/2\right)}\right)^{2}\cdot\frac{n^{2}x^{2}}{4}}$ $=\frac{m^{2}}{n^{2}}$