# How can I calculate this limit without using L'Hopital's rule?

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

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

• A somewhat whimsical comment: Any limit calculation using L'Hôpital can be done without it, but there is a risk that the resulting calculation essentially duplicates the proof of L'Hôpital in a special case. – Harald Hanche-Olsen Oct 21 '14 at 7:20
• Somewhat less whimsically, you can use the Taylor expansions of the numerator and denominator to get the limit. – Harald Hanche-Olsen Oct 21 '14 at 7:21
• Without L'Hôpital, without Taylor, what do you want simpler than Lucian's hints ? – Claude Leibovici Oct 21 '14 at 7:55
• -1 for a title that's totally indistinguishable from the dozens of similar questions shown in the sidebar. – Ilmari Karonen Oct 21 '14 at 13:58

## 5 Answers

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 $f(0)=0$. – Yves Daoust 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. – Yves Daoust Oct 22 '14 at 13:21

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 $\quad\lim_{x\to0}\dfrac{\sin x}x=1$.

• it is $\cos(x)=\pm \sqrt{1-\sin(x)^2}$ – Dr. Sonnhard Graubner Oct 21 '14 at 7:39
• @Dr. Sonnhard Graubner: close to $x=0$, $+$ holds. – Yves Daoust Oct 21 '14 at 7:42
• Thanks, but.. and how do I cancel the $x$ from $\ln{e^x}$ ? – mathsalomon Oct 21 '14 at 17:01
• I can assume that $sin(x) = x$ because of $x\backsimeq0$ ? – mathsalomon Oct 21 '14 at 17:09
• @mathsalomon: $(1)$. The natural logarithm is the inverse of the exponential function. $(2)$. The fact that $\sin x/x\to1$ is independent of l'Hopital, since the proof for the latter theorem relies, either implicitly or explicitly, on the value of the former limit. – Lucian Oct 21 '14 at 17:29

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 $L = 1/2$.

• Why can you discard $\log\cos x/\cos x-1$ ? – Yves Daoust Oct 22 '14 at 7:41
• @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 paramanands.blogspot.com/2014/05/… 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. – Yves Daoust Oct 22 '14 at 13:19
• @YvesDaoust: I updated my answer to make things simpler to understand. – Paramanand Singh Oct 23 '14 at 5:08

yes it is ${\frac {1}{2}}-{\frac {1}{6}}x+{\frac {1}{6}}{x}^{2}-{\frac {7}{360}} {x}^{3}+O \left( {x}^{4} \right)$

• And it can be solved without using power series or Taylor expansion? – mathsalomon Oct 21 '14 at 7:24
• to answer this question needs more time – Dr. Sonnhard Graubner Oct 21 '14 at 7:25

hint：just use Taylor expension, $sinx=x+o(x^{2}),e^{x}=1+x+\frac{x^2}{2!}+o(x^2),cosx=1-\frac{x^2}{2!}+o(x^2).$ $lnx=(x-1)^2-\frac{(x-1)^2}{2}+o(x^2)$. then

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

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