$$\lim_{x \to 0^{+}} (\sin x)^{\cos x}\left(\frac{\cos^{2}x}{\sin x} - \sin x\log(\sin x)\right)$$
The answer is one, but i dont know how to proceed in solving the problem.
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Do you know Taylor expansions of these functions, and what they look like under compositions and exponentiations and multiplications and subtractions of functions, etc.? – user2566092 Aug 06 '15 at 23:11
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No, i am supposed to use only L'Hospital and basic limit calculus techniques – Razvan Paraschiv Aug 06 '15 at 23:12
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Make the substitution $t = \sin x$. – Zhanxiong Aug 06 '15 at 23:13
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ok, then that will make things harder but not impossible. Your main challenge is how to handle the $(\sin x)^{\cos x}$. – user2566092 Aug 06 '15 at 23:14
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Also sin(x)log(sin(x)). – Razvan Paraschiv Aug 06 '15 at 23:17
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$\sin x \approx x , \cos x \approx 1 - \frac{x^2}{2}$ , and $\log x \approx x \log e$ – Hoping_Blessing Aug 06 '15 at 23:20
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@RazvanParaschiv If you can handle $y \log y$ then that's not a problem, you can just work with $\sin x = y$ as $y \to 0$. But you need to express $\cos x$ in terms of $\sin x$ and work that way, similar to another comment suggesting you make this substitution $y = t = \sin x$. – user2566092 Aug 06 '15 at 23:27
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2The limit should be for $x \to 0^{+}$. I have edited your question to fix this. – Paramanand Singh Aug 08 '15 at 11:01
3 Answers
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First note that instead of $x \to 0$ you must have $x \to 0^{+}$ otherwise both $\log\sin x$ and $(\sin x)^{\cos x}$ are undefined. The given expression can be expressed as $$(\sin x)^{\cos x - 1}\cos^{2}x - (\sin x)^{\cos x}\sin x\log\sin x$$ and limit of each term will be evaluated separately.
Now note that as $x \to 0^{+}$, $\cos x \to 1$ and $\log\sin x \to -\infty$ hence $\cos x \log \sin x \to -\infty$ and therefore $$(\sin x)^{\cos x} = \exp(\cos x \log \sin x) \to 0\tag{1}$$ Further note the standard result $$\lim_{t \to 0^{+}}t\log t = 0$$ and hence on putting $t = \sin x$ we get $$\lim_{x \to 0^{+}}\sin x \log \sin x = 0\tag{2}$$ It follows from $(1), (2)$ that $$\lim_{x \to 0^{+}}(\sin x)^{\cos x}\sin x\log \sin x = 0\tag{3}$$ We are now left with the term $$\lim_{x \to 0^{+}}(\sin x)^{\cos x - 1}\cos^{2}x$$ and since $\cos^{2}x \to 1$ we just need to calculate the limit of $(\sin x)^{\cos x - 1}$. This limit will be the final answer. Let it be $L$. Then \begin{align} \log L &= \log\lim_{x \to 0^{+}}(\sin x)^{\cos x - 1}\notag\\ &= \lim_{x \to 0^{+}}\log(\sin x)^{\cos x - 1}\text{ (by continuity of log)}\notag\\ &= \lim_{x \to 0^{+}}(\cos x - 1)\log\sin x\notag\\ &= \lim_{x \to 0^{+}}(\cos x - 1)\log\left(\frac{\sin x}{x}\cdot x\right)\notag\\ &= \lim_{x \to 0^{+}}(\cos x - 1)\log\left(\frac{\sin x}{x}\right) + (\cos x - 1)\log x\notag\\ &= 0\cdot \log 1 + \lim_{x \to 0^{+}}\frac{\cos^{2}x - 1}{\cos x + 1}\cdot\log x\notag\\ &= -\frac{1}{2}\lim_{x \to 0^{+}}\sin^{2}x\log x\notag\\ &= -\frac{1}{2}\lim_{x \to 0^{+}}\frac{\sin^{2}x}{x^{2}}\cdot x\cdot x\log x\notag\\ &= -\frac{1}{2}\cdot 1\cdot 0\cdot 0 = 0\notag \end{align} It follows that $L = e^{0} = 1$.
Paramanand Singh
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Denote the original expression by $I$, and make the substitution $t = \sin x$. Then \begin{align*} & \log I = \sqrt{1 - t^2}\log t + \log\left(\frac{1}{t} - t - t\log t\right) \\ = & \left(1 - \frac{1}{2}t^2 + o(t^2)\right)\log t + \log(1 - t^2 - t^2\log t) - \log t\\ = & \log t - \frac{1}{2}t^2 \log t + \log(1 - t^2 - t^2 \log t) - \log t + o(t^2\log t) \\ = & -\frac{1}{2}t^2\log t - \log(1 - t^2 - t^2\log t) + o(t^2\log t) \\ \to & 0 \end{align*} as $t \to 0$. Hence $I \to 1$ as $t \to 0$.
During the calculation, we used the Taylor expansion of $\sqrt{1 + x}$, and the fact that $\lim\limits_{t \to 0} t^2 \log t = 0$.
Zhanxiong
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$$\lim_{x\to 0 }x^{cos x-1}=1\\\lim_{x\to 0 }\frac{x^{cos x}}{x}=1\\ \xrightarrow[x \to 0]{sin x \approx x} \lim_{x\to 0 }\frac{sin x^{cos x}}{x}=1 $$ name $u(x)=sinx ^{cos x}\\u'(x)=sinx ^{cos x} (\frac {cos^2x}{sinx}-sin x \ln sinx )$ now $$ \lim_{x\to 0 }\frac{sin x^{cos x}}{x}=1\\\lim_{x\to 0 }\frac{u(x)}{x}=\frac{0}{0}=1 $$ l'hosptial because ($\lim_{x\to 0 }sin x^{cos x}=0$ )
$$\lim_{x\to 0 }\frac{u(x)}{x}=\frac{U'(x)}{1}=1 \ \to \\ \lim_{x\to 0 }u'(x)=1 \\\lim_{x\to 0 }u'(x) =\lim_{x\to 0 }sinx ^{cos x}\\u'(x)=sinx ^{cos x} (\frac {cos^2x}{sinx}-sin x \ln sinx )=1 $$
Khosrotash
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1Apart from the initial non-rigorous steps you have made a very serious mistake in applying LHR (and hence the downvote from me). You first show that $u(x)/x \to 1$ and then say that after LHR $u'(x)/1$ also tends to same limit. This should be called "LHR in reverse" and it is wrong. LHR says that if $f'(x)/g'(x) \to L$ then under certain conditions $f(x)/g(x) \to L$. It does not say that if $f(x)/g(x) \to L$ then $f'(x)/g'(x) \to L$. – Paramanand Singh Aug 08 '15 at 08:26