Question

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Answer 1

`\displaystyle \lim_(x\to 0)\left((2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3)/(1-\cos \left(3x\right))\right)`

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

`\displaystyle \lim_(x\to 0)\left(2\ln(1+3x)+(6x)/(1+3x)+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)`

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

`\displaystyle \lim_(x\to0)(2\ln(1+3x))/(3\sin(3x)) + \lim_(x\to0)(6x)/(3(1+3x)\sin(3x)) \\\\ + \lim_(x\to0)(\cos(x)\tan(3x))/(3\sin(3x)) + \lim_(x\to0)(3\sin(x)\sec^2(x))/(3\sin(3x)) \\\\ - \lim_(x\to0)(6x^2)/(3\sin(3x))`

Now recall two well-known limits:

`\displaystyle \lim_(x\to0)(\sin(ax))/(ax)=1\text{ if }a\\eq0 \\\\ \lim_(x\to0)(\ln(1+ax))/(ax)=1\text{ if }a\\eq0`

Compute each remaining limit:

`\displaystyle \lim_(x\to0)(2\ln(1+3x))/(3\sin(3x)) = \frac23 * \lim_(x\to0)(\ln(1+3x))/(3x) * \lim_(x\to0)(3x)/(\sin(3x)) = \frac23`

`\displaystyle \lim_(x\to0)(6x)/(3(1+3x)\sin(3x)) = \frac23 * \lim_(x\to0)(3x)/(\sin(3x)) * \lim_(x\to0)(1)/(1+3x) = \frac23`

`\displaystyle \lim_(x\to0)(\cos(x)\tan(3x))/(3\sin(3x)) = \frac13 * \lim_(x\to0)(\cos(x))/(\cos(3x)) = \frac13`

`\displaystyle \lim_(x\to0)(3\sin(x)\sec^2(x))/(3\sin(3x)) = \frac13 * \lim_(x\to0)\frac{\sin(x)}x * \lim_(x\to0)(3x)/(\sin(3x)) * \lim_(x\to0)\sec^2(x) = \frac13`

`\displaystyle \lim_(x\to0)(6x^2)/(3\sin(3x)) = \frac23 * \lim_(x\to0)x * \lim_(x\to0)(3x)/(\sin(3x)) * \lim_(x\to0)x = 0`

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2