Roger Smith

2021-12-30

$\underset{x\to 0}{lim}\frac{\mathrm{sin}h-\mathrm{sin}x}{x\left(\mathrm{cos}h\left(x\right)-\mathrm{cos}\left(x\right)\right)}$

Buck Henry

Hint: In fact,
$L=\underset{x\to 0}{lim}\frac{\mathrm{sin}h\left(x\right)-\mathrm{sin}\left(x\right)}{x\left(\mathrm{cos}h\left(x\right)-\mathrm{cos}\left(x\right)\right)}$
$=\underset{x\to 0}{lim}\frac{\mathrm{sin}h\left(x\right)-\mathrm{sin}\left(x\right)}{{x}^{3}}\cdot \underset{x\to 0}{lim}\frac{{x}^{2}}{\mathrm{cos}h\left(x\right)-\mathrm{cos}\left(x\right)}$
$=\underset{x\to 0}{lim}\frac{\mathrm{cos}h\left(x\right)-\mathrm{cos}\left(x\right)}{3{x}^{2}}\cdot \underset{x\to 0}{lim}\frac{2x}{\mathrm{sin}h\left(x\right)+\mathrm{sin}\left(x\right)}$

Dabanka4v

Since $\mathrm{sin}hx-\mathrm{sin}x\sim \frac{1}{3}{x}^{3}$ while $\mathrm{cos}hx-\mathrm{cos}x\sim {x}^{2}$, the limit is $\frac{1}{3}$

Vasquez

$\begin{array}{}\underset{x\to 0}{lim}\frac{\mathrm{sin}h\left(x\right)-\mathrm{sin}\left(x\right)}{x\left(\mathrm{cos}h\left(x\right)-\mathrm{cos}\left(x\right)\right)}\\ =\underset{x\to 0}{lim}\frac{\left(x+\frac{{x}^{6}}{6}+o\left({x}^{4}\right)\right)-\left(x-\frac{{x}^{3}}{6}+o\left({x}^{4}\right)\right)}{x\left(\left(1+\frac{{x}^{2}}{2}+o\left({x}^{3}\right)\right)\right)-\left(1-\frac{{x}^{2}}{2}+o\left({x}^{3}\right)\right)}\\ =\underset{x\to 0}{lim}\frac{\frac{{x}^{3}}{3}+o\left({x}^{4}\right)}{x+o\left({x}^{4}\right)}\\ =\underset{x\to 0}{lim}\frac{\frac{1}{3}+o\left({x}^{3}\right)}{1+o\left({x}^{3}\right)}\\ =\frac{1}{3}\end{array}$