Question

Find the limit, if it exists, analytically using special trig limits

Answer 1

`\displaystyle \lim_(x\to 0)~\cfrac{1-cos^2(7x)}{8x^2(~~1-sin^2(7x)~~)}~\hfill \boxed{u=7x\implies \cfrac{u}{7}=x\implies \cfrac{u^2}{49}=x^2} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1-cos^2(u)}{8\left((u^2)/(49) \right)(~~1-sin^2(u)~~)}\implies \cfrac{1-cos^2(u)}{ ~~ (8u^2(~~1-sin^2(u)~~))/(49) ~~ }\implies \cfrac{[1-cos^2(u)]49}{8u^2(~~1-sin^2(u)~~)} \\\\\\ \cfrac{[1-cos(u)][1-cos(u)]49}{8u^2(~~1-sin^2(u)~~)}\implies \cfrac{1-cos(u)}{u}\cdot \cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]} \\\\[-0.35em] ~\dotfill`

`\displaystyle \lim_(x\to 0)~\cfrac{1-cos(u)}{u}\cdot \cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]}\implies \lim_(x\to 0)~\cfrac{1-cos(u)}{u}\cdot \lim_(x\to 0)~\cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]} \\\\\\ \displaystyle \lim_(x\to 0)~\cfrac{1-cos(7x)}{7x}\cdot \lim_(x\to 0)~\cfrac{49[1-cos(7x)]}{8(7x)[1-sin^2(7x)]} \\\\\\ \displaystyle \text{\LARGE 0}\cdot \lim_(x\to 0)~\cfrac{49[1-cos(7x)]}{8(7x)[1-sin^2(7x)]}\implies \text{\LARGE 0}`