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Question Number 192433 by cortano12 last updated on 18/May/23

$$\underset{x\to 0}{\lim }\frac{1-\cos \left(\ln (3\mathrm{x}+1)\right)}{{3\mathrm{x}}^6-\tan \left({3\mathrm{x}}^2\right)}=?$$

Answered by mehdee42 last updated on 18/May/23

$$tip:ifu\to 0$$$$\Rightarrow 1-cosu\sim \frac{1}{2}{u}^2\mathrm{\&}ln(u+1)\sim u\mathrm{\&}tanu\sim u$$$$\Rightarrow {lim}_{x\to 0}\frac{1-co\left(ln(3x+1)\right)}{3x^6-tan\left(3x^2\right)}={lim}_{x\to 0}\frac{\frac{1}{2}{\left(ln(3x+1)\right)}^2}{3x^6-3x^2}={lim}_{x\to 0}\frac{\frac{9x^2}{2}}{3x^2(x^4-1)}=-\frac{3}{2}✓$$

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