rhedynogh0rp

2022-03-30

I am trying to find an easy way to compute the limit as $x\to 0$ of
$f\left(x\right)=\frac{\sqrt{1+\mathrm{tan}\left(x\right)}-\sqrt{1+\mathrm{sin}\left(x\right)}}{{x}^{3}}$

Mercedes Chang

$\frac{1-\mathrm{cos}\left(x\right)}{{x}^{2}}$
$=\frac{2{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}{{x}^{2}}$
For x small enough , we have:
$\mathrm{tan}x>x\mathrm{sin}x$
(using the geometric interpretation)
then
$\mathrm{cos}\left(x\right)<\frac{\mathrm{sin}x}{x}<1$
and since the function $\mathrm{cos}\left(x\right)$ is a continuous function
$\underset{x\to 0}{lim}\mathrm{cos}x=\mathrm{cos}0=1$
apply this to above, also recall squeeze theorem, we get what you want.
How to prove $\mathrm{cos}x$ is continunous, you may ask.
$|\mathrm{cos}x-\mathrm{cos}y|=2|\mathrm{sin}\left(\frac{x+y}{2}\right)\mathrm{sin}\left(\frac{x-y}{2}\right)|\le 2|\mathrm{sin}\left(\frac{x-y}{2}\right)|$
now we only have to prove
$\underset{x\to 0}{lim}\mathrm{sin}x=0$
for x small enough,
$0<|\mathrm{sin}x|<|x|$
then
$0\le \underset{x\to 0}{lim}\mathrm{sin}x|\le \underset{x\to 0}{lim}|x\mid =0$

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