Evaluating $\lim\limits_{x\to 0} \frac{\ln(1+\sqrt[3]{x})}{\sin^4(\sqrt[3]{x})}$ I am kind of confused trying to compute  $$ \lim_{x\to 0}  \frac{\ln(1+\sqrt[3]{x})}{\sin^4(\sqrt[3]{x})}. $$
I got that it is of the form $(\frac {0}{0})$, since  $\ln(1+x)=x $ so it is  $ \ln(1+\sqrt[3]{x})= \sqrt[3]{x}  $, and $\sin x=x $ so   $ \sin^4(\sqrt[3]{x}) $ =$\sqrt[3]{x}$, so the limit becomes 
$\lim_{x\to 0}  \frac{\sqrt[3]{x}}{(\sqrt[3]{x})^4} $.
Am I right this far?
Still I don't know what to do next because I still get 0 as the answer.
Any suggestions?
 A: I'm going to make a small substitution to make this a bit easier, $u=x^{1/3}$.  Conveniently, we still take the limit toward zero, $ \lim_{u \to 0} \frac{ \ln{ (1 + u) }}{\sin^4 (u)}$.
If we evaluate the expression at $u=0$, it's still indeterminate, but a little less messy.  If you've used $\epsilon - \delta$ definitions for the limit, you know the trick is to go away from the limiting value (zero in this case) just a bit.  Maybe to $0.1$ and $-0.1$.  For $u = 0.1$ the expression evaluates to $959.5$.  You should be thinking, "Hm, maybe this thing tends toward infinity."  For $u = -0.1$, however, the expression evaluates to $-1060.7$.  That's sounding like negative infinity.  Sure, it's possible that this expression just has really big numbers and will eventually converge toward a value at zero, but it's unlikely.  More likely, the limits from the left and the from the right don't converge, i.e., $\lim_{u \to 0 +} \neq \lim_{u \to 0 -}$, in which case the two sided limit $\lim_{u \to 0}$ is undefined.  That is, in fact, the case, but I'll prove it.
I will use a first order approximation for the numerator and denominator, $f(x) \approx f(0) + \left. \frac{df}{dx} \right|_{x=0} \Delta x $.  This can be made arbitrarily accurate simply by choosing smaller and smaller $\Delta x$.  With the $\epsilon - \delta$ definition of the limit, that's exactly what we're doing.  So, for our two expressions,
$$
\ln (1 + u ) \approx \ln (1 + 0 ) + \left. \frac{1}{1+u} \right|_{u = 0} \Delta u = \Delta u
\\
\sin^4 (u) \approx \sin^4 (0) + \left. 4 \sin^3 (u) \cos{(u)} \right|_{u=0} \Delta u = 0
$$ 
From this, we see that if we go slightly to the right of $u=0$, for example, $u=0.1$, the numerator will be a small positive number, and the denominator will be zero.  Well, a positive number divided by zero is $\infty$.  However, if we go slightly left, for example, $u = -0.1$, we see that the numerator of the expression will be a small negative number divided by zero.  That limits to $-\infty$.  The result taken together is
$$
\lim_{u \to 0+} \frac{ \ln{ (1 + u) }}{\sin^4 (u)} = +\infty
\\
\lim_{u \to 0-} \frac{ \ln{ (1 + u) }}{\sin^4 (u)} = -\infty
$$
Because the right and left limits do not converge, the two sided limit is undefined.
A: the limit does not exist.because$\lim_{x\to0+}$ is $\frac{+ve}{+ve}=+ve$ and $\lim_{x\to0-}$ is $\frac{-ve}{+ve}=-ve$
