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$$\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$$

Can this limit be calculated without using L'Hopital's rule?

Saurabh Rana
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mathsalomon
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5 Answers5

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Hints: $\quad\ln(ab)=\ln a+\ln b,\qquad\ln e^x=x,\qquad\cos x=\sqrt{1-\sin^2x},\qquad\ln a^b=b\ln a,$

$\ln(1+t)\simeq t\quad$ when $\quad t\simeq0,\quad$ and $\quad\lim_{x\to0}\dfrac{\sin x}x=1$.

Lucian
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Clearly $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin x - \log(e^{x}\cos x)}{x\sin x}\\ &= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\cdot\frac{x}{\sin x}\\ &= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\sin x - x}{x^{2}} - \frac{\log \cos x}{x^{2}}\\ &= A - B\end{aligned}$$ Next we can see that $$A = 0$$ from here. And we have $$\begin{aligned}B &= \lim_{x \to 0}\frac{\log\cos x}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\frac{\cos x - 1}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\lim_{x\to 0}\frac{\cos x - 1}{x^{2}}\\ &= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x\to 0}\frac{\cos x - 1}{x^{2}}\text{ (putting } t = \cos x - 1)\\ &= \lim_{x \to 0}\frac{\cos x - 1}{x^{2}}\\ &= \lim_{x \to 0}\frac{\cos^{2} x - 1}{x^{2}(1 + \cos x)}\\ &= -\frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = -\frac{1}{2}\end{aligned}$$ It follows that $L = 1/2$.

Community
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Paramanand Singh
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  • Why can you discard $\log\cos x/\cos x-1$ ? –  Oct 22 '14 at 07:41
  • @YvesDaoust: when $x\to 0$ then $y=\cos x\to 1$ and then $(\log y)/(y - 1) \to 1$ as $y \to 1$. This last limit is fundamental for logarithmic functions and can be proved without l'Hospital using any definition of logarithm function. See my blog series http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-1.html for various definitions of $\log$ and $\exp$ and proofs for such fundamental limits. – Paramanand Singh Oct 22 '14 at 10:48
  • For your answer to be complete, you should indicate these intermediate steps so that the reader doesn't have to guess them. –  Oct 22 '14 at 13:19
  • @YvesDaoust: I updated my answer to make things simpler to understand. – Paramanand Singh Oct 23 '14 at 05:08
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Rewriting the numerator as $\sin x-x-\ln\cos x$, notice that $\sin x-x$ is an odd function; as the denominator is even, the ratio vanishes at $x=0$ and these terms can be ignored.

Then $$\lim_{x\to0}-\frac{\ln\cos x}{x\sin x}=-\lim_{x\to0}\frac{\frac12\ln(1-\sin^2x)}{\sin^2x}\frac{\sin x}x=-\frac12\lim_{t\to0^+}\frac{\ln(1-t)}t=-\frac12\lim_{t\to0^+}\ln(1-t)^{1/t}\\=-\frac12\ln\left(\lim_{t\to0^+}(1-t)^{1/t}\right)=-\frac12\ln e^{-1}=\frac12.$$

UPDATE:

We need to show that the first limit exists. Without being allowed to use derivatives, we need some property of the trigonometric functions, and we admit $\sin x\le x\le \tan x$, so that $$1\ge\frac{\sin x}x\ge\cos x\ge\cos^2x,$$ and from there $$0\ge\frac{\sin x-x}x\ge\cos^2x-1,$$ $$0\ge\frac{\sin x-x}{x\sin x}\ge-\sin x.$$ As expected, the limit is $0$. As a byproduct, the same relations establish the limit of $\sin x/x$.

  • many thanks. And.. What is the relationship made ​​it 'odd-even' with the annulment? – mathsalomon Oct 21 '14 at 17:49
  • Any function can be decomposed in an odd and even part. The odd part, such that $f(-x)=-f(x)$ is implicitly such that $f(0)=0$. –  Oct 21 '14 at 18:15
  • Note that $g(x)=(\sin x - x)/(x\sin x)$ is odd but not defined at $x=0$. But being odd does not guarantee that the limit is $0$. As counterexample consider $f(x)=\sin(1/x)$ which is odd but does not tend to any limit as $x\to 0$. However, there is a theorem which says that if $f(x)$ is odd and $\lim_{x\to 0}f(x)$ exists then this limit must be $0$. Note that existence of the limit is a prerequisite for this theorem to work. Hence your solution must show that $\lim\limits_{x\to 0}\dfrac{\sin x - x}{x\sin x}$ exists. – Paramanand Singh Oct 22 '14 at 10:59
  • @Paramanand Singh: I agree with you, the limit must exist. –  Oct 22 '14 at 13:21
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yes it is ${\frac {1}{2}}-{\frac {1}{6}}x+{\frac {1}{6}}{x}^{2}-{\frac {7}{360}} {x}^{3}+O \left( {x}^{4} \right) $

Dr. Sonnhard Graubner
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Hint: Just use Taylor expension, $\sin x=x+o(x^{2}),\;\cos x=1-\frac{x^2}{2!}+o(x^2),\;$ $\ln(1+x)=x-\frac{x^2}{2}+o(x^2).$ Then

$\ln(e^x\cos x)=x+\ln\cos x=x-\frac{x^2}{2}+o(x^2)$. Hence

$\lim_{x\to 0} \frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}} = \lim_{x\to0} \frac{x-(x-\frac{x^2}{2})+o(x^2)}{x^2} =\frac{1}{2} .$

Jacob.Lee
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