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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?

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No, you are not right. None of those equal signs make any sense, to begin with. $\ln(1+x)$ is definitely not equal to $x$, etc. – Andrés E. Caicedo Nov 12 '13 at 18:14
Please, make an effort to write in proper English and not "text speak". I wud b g8ful! – Fly by Night Nov 12 '13 at 18:15
@FlybyNight These were relatively mild offences... No "u", "pls" or "cuz" in sight. The title choice was far worse. – Lord_Farin Nov 12 '13 at 18:19
@Lord_Farin Did you see the original post? A mass of "..." and "???" along with lots of "i" in place of "I". A lazy, slapdash post shows lack of thought, lack of consideration and, ultimately, lack of respect for the recipient; that's you and me! – Fly by Night Nov 12 '13 at 18:21
@FlybyNight Yes. That's not text speak. That's laziness/habit. Not nice, I agree. But not on the scale of all-caps or actual text speak. It's not necessarily lack of respect; it could just be lack of consideration, or the OP's first contribution to a web site that propagates proper English. – Lord_Farin Nov 12 '13 at 18:29
up vote 1 down vote accepted

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.

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the limit does not exist.because$\lim_{x\to0+}$ is $\frac{+ve}{+ve}=+ve$ and $\lim_{x\to0-}$ is $\frac{-ve}{+ve}=-ve$

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Existence of a limit is based on equality of left and right hand limits, but just by judging sign of the expression and noting that it is different from left and right does not lead to the fact that limit does not exist. If left hand and right hand limits were zero then the limit would exist even when the signs are different. Here the existence fails because one limit is $\infty$ and other is $-\infty$ – Paramanand Singh Nov 13 '13 at 5:42

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