Santino Bautista

2022-06-30

I used all the trig and log tricks but still can't compute this limit
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+{\mathrm{sin}}^{2}\left(2x\right)\right)}{1-{\mathrm{cos}}^{2}\left(x\right)}.$

Bornejecbo

Expert

Just using the standard limit $\underset{t\to 0}{lim}\frac{\mathrm{ln}\left(1+t\right)}{t}=1$ you get
$\begin{array}{rcl}\frac{\mathrm{ln}\left(1+{\mathrm{sin}}^{2}\left(2x\right)\right)}{1-{\mathrm{cos}}^{2}\left(x\right)}& =& \frac{\mathrm{ln}\left(1+{\mathrm{sin}}^{2}\left(2x\right)\right)}{{\mathrm{sin}}^{2}\left(2x\right)}\cdot \frac{{\mathrm{sin}}^{2}\left(2x\right)}{{\mathrm{sin}}^{2}x}\\ & =& 4{\mathrm{cos}}^{2}x\cdot \frac{\mathrm{ln}\left(1+{\mathrm{sin}}^{2}\left(2x\right)\right)}{{\mathrm{sin}}^{2}\left(2x\right)}\\ & \stackrel{x\to 0}{⟶}& 4\cdot 1\cdot 1=4\end{array}$

gnatopoditw

Expert

We know that (by asymptotic relations):
$\mathrm{ln}\left(1+f\left(x\right)\right)\sim f\left(x\right)\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x\to \alpha \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)\to 0$
Also:
$\mathrm{sin}\left(x\right)\sim t\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x\to 0$
And:
$\mathrm{cos}\left(x\right)\sim 1-\frac{1}{2}{x}^{2}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x\to 0$
So, your limit is going to be:
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+{\mathrm{sin}}^{2}\left(2x\right)\right)}{1-{\mathrm{cos}}^{2}\left(x\right)}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sim \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\underset{x\to 0}{lim}\frac{{\mathrm{sin}}^{2}\left(2x\right)}{1-{\mathrm{cos}}^{2}\left(x\right)}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\underset{x\to 0}{lim}\frac{4{x}^{2}}{1-{\mathrm{cos}}^{2}\left(x\right)}=\underset{x\to 0}{lim}\frac{4{x}^{2}}{\left(1+\mathrm{cos}\left(x\right)\right)\cdot \left(1-\mathrm{cos}\left(x\right)\right)}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\underset{x\to 0}{lim}\frac{4{x}^{2}}{2\cdot \frac{1}{2}{x}^{2}}=4$

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