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I want to calculate : $$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$ with l'Hôpital's rules. I get $$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right)$$ but I don't see useful next steps.

Guy Fsone
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Anna Saabel
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7 Answers7

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Since the question has been tagged as "real analysis", I assume that the OP is familiar with Taylor series: The idea of using L'hopital's rule for trigonometric functions is generally not good because when you differentiate a trigonometric function, you again get a new trigonometric function. + differentiation of the multiplication of two terms (in the denominator) makes it more complicated. So, L'hopital is not very efficient in this problem.

You can do it more easily using approximates of the Taylor series for $\sin(x)$:

$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)=\lim\limits_{x\to 0}\frac{x-\sin(x)}{x\sin(x)}=\lim\limits_{x\to 0}\frac{x^3/6}{x\sin(x)}=\lim\limits_{x\to 0}x/6\frac{x}{\sin(x)}=0\times1=0$$

Here's an alternative way using only L'hopital's rule:

$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)=\lim\limits_{x\to 0}\frac{x-\sin(x)}{x\sin(x)}=\lim\limits_{x\to 0}\frac{1-\cos(x)}{\sin(x)+x\cos(x)}=\lim\limits_{x\to 0}\frac{\frac{1-\cos(x)}{x}}{\frac{\sin(x)}{x}+\cos(x)}=\lim\limits_{x\to 0}\frac{1-\cos(x)}{x}\times \lim\limits_{x\to 0}\left(\frac{1}{\frac{\sin(x)}{x}+\cos(x)}\right)=0\times \frac{1}{2}=0$$

Use L'hopitals rule separately to prove that:

$$\lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$$ $$\lim\limits_{x\to 0}\frac{1-\cos(x)}{x}=0$$

stressed out
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  • +1 for the advice in first paragraph. IMHO it is always preferable to use Taylor instead of the infamous L'Hospital's Rule. – Paramanand Singh Dec 12 '17 at 15:27
  • @ParamanandSingh: Thanks. I totally agree with you about the preference of Taylor series over L'hopital's rule in most cases that we deal with in Calculus. I mean, when a function is analytic, certainly Taylor series is preferred over L'hopital's rule. However, L'hopital's rule has wider applicability because it requires only differentiability in an open neighborhood containing the point of evaluation where the derivative of the denominator is not $0$ on the interval. Therefore, applying L'hopital's rule a finite number of times to evaluate an indeterminate form has much weaker assumptions. – stressed out Dec 12 '17 at 15:52
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Not taking derivative again, we can also look that $1-\cos x = 2 \sin^2 (x/2)$. Using this you get zero as limit.

$$\lim\limits_{x \rightarrow 0}\left(\frac{\tfrac{2\sin^2(x/2)}{x}}{\tfrac{\sin(x)}{x}+\cos(x)}\right) = 0$$

gilly
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  • But why you see that is 0 ? For example $\frac{sin(x)}{x}=\frac{0}{0} $ ..? – Anna Saabel Dec 12 '17 at 13:41
  • @AnnaSaabel: It is a well-known fact that $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$. This, you can prove easily using L'hopital's rule. But there are many more proofs, even elementary non-algebraic/not-very-rigorous proofs that use geometry. – stressed out Dec 12 '17 at 13:43
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You can also use Taylor series: $$\sin(x)=x-\frac{x^3}{6}+O(x^4)$$ So your function: $$\frac{1}{\sin(x)}-\frac{1}{x}=\frac{x-\sin(x)}{x\sin(x)}=\frac{\frac{x^3}{6}+O(x^4)}{x^2-\frac{x^4}{6}+O(x^5)}=\frac{\frac{x}{6}+O(x^2)}{1-\frac{x^2}{6}+O(x^3)} \to \frac{0}{1}=0$$

Botond
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just use it again

$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) =\\ \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{\sin(x)}{\cos(x)-x\sin(x)+\cos(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{0}{1-0\cdot0+1}\right)=0$$

ℋolo
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As it turns out, you just have to use L'Hospital's rule twice and observe that on the second try the denominator is no longer zero when you use directly substitution:

\begin{align} \lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin x} - \frac{1}{x}\right) &=\lim\limits_{x \rightarrow 0}\frac{x -\sin x}{x\sin x}\\ &=\lim\limits_{x \rightarrow 0}\frac{(x -\sin x)'}{(x\sin x)'}\\ &=\lim\limits_{x \rightarrow 0}\frac{x' -(\sin x)'}{x'\sin x+x(\sin x)'}\\ &=\lim\limits_{x \rightarrow 0}\frac{1 - \cos x}{\sin x+x \cos x}\\ &=\lim\limits_{x \rightarrow 0}\frac{(1 - \cos x)'}{(\sin x+x \cos x)'}\\ &=\lim\limits_{x \rightarrow 0}\frac{1' - (\cos x)'}{(\sin x)'+x' \cos x +x(\cos x)'}\\ &=\lim\limits_{x \rightarrow 0}\frac{\sin x}{\cos x + \cos x - x \sin x}\\ &=\lim\limits_{x \rightarrow 0}\frac{\sin x}{2\cos x - x \sin x}\\ &=\frac{0}{2\cdot 1 - 0 \cdot 0}\\ &=\frac{0}{2}\\ &=0 \end{align}

Michael Rybkin
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If you are forced to use the infamous L'Hospital's Rule then you should first try to simplify the expression and then apply L'Hospital's Rule.

We can proceed as follows \begin{align} L&=\lim_{x\to 0}\frac{1}{\sin x} - \frac{1}{x}\notag\\ &=\lim_{x\to 0}\frac{x-\sin x} {x\sin x} \notag\\ &=\lim_{x\to 0}\frac{x-\sin x} {x^{2}}\cdot\frac{x}{\sin x} \notag\\ &=\lim_{x\to 0}\frac{x-\sin x} {x^{2}} \notag \\ &=\lim_{x\to 0}\frac{1-\cos x} {2x}\text{ (via L'Hospital's Rule)} \notag \\ &=\frac{1}{2}\lim_{x\to 0}\frac{1-\cos^{2}x}{x(1+\cos x)} \notag\\ &=\frac{1}{4}\lim_{x\to 0}x\cdot\left(\frac{\sin x} {x} \right) ^{2}\notag\\ &=\frac{1}{4}\cdot 0\cdot 1^{2}\notag \\ &=0\notag \end{align}

Paramanand Singh
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Sum the fraction Then apply easily L'Hopiyal rule twice

$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\left(\frac{\sin x - x}{x\sin x}\right) \\= \lim_{x \rightarrow 0}\left(\frac{\cos x - 1}{\sin x +x\cos x}\right)=\lim_{x \rightarrow 0}\left(\frac{-\sin x }{2\cos x -x\sin x }\right)=0$$

Guy Fsone
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