I am struggling to make this limit: $$\lim_{x\to 0} \frac{\log(1+\sqrt[3]{x})(\sin x - x)}{(\cos x-1)^3x^2}$$ I know it doesn't exist, and I am really near to prove it but something is blocking me. Skipping some step, I have that $$\frac{\sqrt[3]x(\sin x-x)}{-\frac{1}{2}x^8}$$ From this, I see that $x^8$ could be $x^7$, by making the top term (1-1), and from this I see that this goes to +-infinity, so it doesn't exist. But this isn't a formal method, nor probably a method at all. Would love to see your methods to solve this without hopital!
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1See http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Nov 18 '16 at 17:31
4 Answers
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Just rewrite it as
$$\frac{\ln (1+\sqrt[3]{x})}{\sqrt[3]{x}}\sqrt[3]{x}\frac{\sin x -x}{x^3}x^3\frac{x^6}{(\cos x-1)^3}\frac{1}{x^6}\frac{1}{x^2}$$ you get a limit of $\infty$.
Rene Schipperus
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Sorry, I don't get how to deal with $(sinx-x)$, neither in your way – L. Repetti Nov 18 '16 at 17:08
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@L.Repetti (sinx)/x tends to 1 as x tends to 0, have you covered this identity? – Hugh Nov 18 '16 at 17:12
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Thanks for the solution and for the explanation, but I don't get why; is this a known limit, or should I use some properties? EDIT: Is it because sin x is smaller than x so we can "trow it away" and keep x? – L. Repetti Nov 18 '16 at 17:15
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@alans yeah I know I couldnt edit my comment glad you cleared it up. – Rene Schipperus Nov 18 '16 at 18:39
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Use Taylor expansion of $\sin(x)$ about $0$ which is
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} ...$
with which you'll get
$\frac{x^{\frac{3}{2}}(- \frac{x^3}{3!} + \frac{x^5}{5!} ...)}{\frac{-1}{2}x^8}$. You can follow from here.
Dhruv Kohli
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We still don't "know" Taylor at university, and we can use just what we have done so I should figure out another way; anyway, thanks! – L. Repetti Nov 18 '16 at 17:10
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For all $\varepsilon>0$ and for all $x$ sufficiently close to $0$ we have: $$ \begin{align*} (1-\varepsilon)\sqrt[3]{x}&\le\ln(1+\sqrt[3]{x})\\ -\varepsilon x&\le\sin(x)-x\\ \varepsilon x^2&\ge\cos(x)-1. \end{align*} $$ Thus for $c=\frac{\varepsilon^2-\varepsilon}{\varepsilon^3}$ we have $$ \lim_{x\to0}\frac{c}{x^{20/3}}=\lim_{x\to0}\frac{((1-\varepsilon)\sqrt[3]{x})(-\varepsilon x)}{(\varepsilon x^2)^3x^2}\le\lim_{x\to0}\frac{\ln(1+\sqrt[3]{x})(\sin(x)-x))}{(\cos(x)-1)^3x^2}. $$ It is easily seen that the LHS diverges as $x\to 0$ which forces the RHS to diverge as well.
Laars Helenius
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You opened my mind to a new world that I already knew but now, with your way to solve, I really feel as true. I'm used to < in series, but in limits the squeeze way was something so "external". Really, this method is the fastest. – L. Repetti Nov 18 '16 at 17:55
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A general comment...
Even though you are not allowed to use l'H$\hat{\operatorname{o}}$pital's, I do believe you may use things like the definition of derivative, product and quotient rule for limits .
For example $$\lim_{x\to0}\frac{\log(1+x)}{\sin x} =\lim_{x\to0}\frac{\frac{\log(1+x)-\log(0)}{x}}{\frac{\sin x-\sin 0}{x}} =\frac{\frac{d}{dx}\log(1+x)\big|_{x=0}}{\frac{d}{dx}\sin x\big|_{x=0}}=\frac{1/(1+0)}{\cos 0}=1$$
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We can use just we have seen / demostrated, and (in university) we haven't a clue about what derivatives are, so unfortunately we can't use them! – L. Repetti Nov 19 '16 at 10:28