# How to find $\lim_{x \to 0} ( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)})$

I want to do the problem without using L'Hopitals rule, I have $$\frac{1}{\sin(x)}- \frac{1}{\arcsin(x)} = \frac{x}{\sin(x)}\frac{x}{\arcsin(x)}\frac{\sin(x)-\arcsin(x)}{x^2}$$ and I'm not quite sure about how to deal with the $\dfrac{\sin(x)-\arcsin(x)}{x^2}$, apparently its limit is $0$? In which case the whole limit would be $0$. But how would I show this without using l'Hopitals rule. Thanks for any help.

• If you can't use L'Hopital, you can use the power series for $\sin x$ and $\arcsin x$... – Thomas Andrews Sep 6 '14 at 12:39
• Without L'Hopital's rule and without the full power of Taylor series, you can note that $\sin(0)=\arcsin(0)=0$ (so the numerator is $o(1)$) and $\sin'(0)=\arcsin'(0)=1$ (so the numerator is $o(x)$). Now $\sin''(0)=0$ is clear, so all that's left to be done is checking that $\arcsin''(0)=0$ to conclude that the numerator is $o(x^2)$ and so the whole thing goes to 0. – Ian Sep 6 '14 at 13:40
• @Ian: The whole point of not using L'Hospital and Taylor is to avoid differentiation and instead find a solution using rules of limits and some manipulation of the expression under consideration. I have presented an answer along these lines. – Paramanand Singh Sep 6 '14 at 18:22

Using Taylor's theorem: $$\color{blue}{\sin x=x-x^3/6+x^5/120+O(x^7)}$$ $$\color{red}{\arcsin x=x+x^3/6+3x^5/40+O(x^7)}$$ So: \begin{align}\left( \frac{1}{\color{blue}{\sin(x)}}- \frac{1}{\color{red}{\arcsin(x)}}\right)&=\frac{\color{red}{\arcsin x}-\color{blue}{\sin x} }{\color{blue}{\sin x}\arcsin x}\\&=\frac{\color{blue}{\color{red}{(x+x^3/6+O(x^5))}-(x-x^3/6+O(x^5))}}{\color{blue}{(x-x^3/6+O(x^5))}\color{red}{(x+x^3/6+O(x^5))}}\\&=\frac{+x^3/3+O(x^5)}{x^2+O(x^6)}\end{align} So: $$\lim \limits_{x \to 0}\left( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)}\right)=0$$

When $x\rightarrow0$

• $\sin x\sim x-\frac{x^3}{6}$
• $\arcsin x\sim x+\frac{x^3}{6}$
• $\sin x-\arcsin x \sim (x-\frac{x^3}{6})-(x+\frac{x^3}{6})=-\frac{x^3}{3}$

$$\displaystyle \lim_{x\rightarrow 0}\frac{\sin x -\arcsin x}{\sin x \cdot \arcsin x }=\lim_{x\rightarrow 0}\frac{-\frac{x^3}{3}}{\sin x \cdot \arcsin x }=\lim_{x\rightarrow 0}\frac{-\frac{x^3}{3}}{ x ^2 }=\lim_{x\rightarrow 0}\frac{\frac{-x}{3}}{ 1 }=0\\$$

Let $x \to 0^{+}$ and then we know that $\cos x < \dfrac{\sin x}{x} < 1$ (this is pretty standard from geometric arguments commonly used in proof of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ see this answer) and then we get $$\cos x - 1 < \frac{\sin x - x}{x} < 0$$ which further means that $$\frac{\cos x - 1}{x} < \frac{\sin x - x}{x^{2}} < 0\tag{1}$$ Now we can see that $$\frac{\cos x - 1}{x} = -\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x} \to 0$$ as $x \to 0^{+}$. Hence by using squeeze theorem on $(1)$ we get $$\lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} = 0$$ and hence \begin{aligned}\lim_{x \to 0^{+}}\frac{\arcsin x - x}{x^{2}} &= \lim_{x \to 0^{+}}\frac{\arcsin x - x}{(\arcsin x)^{2}}\cdot\frac{(\arcsin x)^{2}}{x^{2}}\\ &= \lim_{t \to 0^{+}}\frac{t - \sin t}{t^{2}}\cdot\frac{t^{2}}{\sin^{2}t}\text{ (}\arcsin x = t)\\ &= 0\end{aligned} It is now clear that $$\lim_{x \to 0^{+}}\frac{\sin x - \arcsin x}{x^{2}} = \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} - \frac{\arcsin x - x}{x^{2}} = 0$$ If $x \to 0^{-}$ then we can put $x = -y$ and $y \to 0^{+}$ and again the limit is $0$.

This is a nice example which shows that most usual limit problems don't need advanced tools like L'Hospital's Rule and Taylor series.

• you showed $L_1=\frac{\sin x-x}{x^2}\to0$ and $L_2=\frac{\arcsin x-x}{x^2}\to0$ when $x\to0$ in both but that doesn't mean $L_1-L_2\to0$ – RE60K Sep 9 '14 at 11:08
• @Aditya: Your comment is weird and i find hard to believe it is coming from you. If $f(x) \to A$ and $g(x) \to B$ as $x \to a$ then $f(x) - g(x) \to (A - B)$ as $x \to a$. Here $a = A = B = 0$ and $f(x) = (\sin x - x)/x^{2}$ and $g(x) = (\arcsin x - x)/x^{2}$. – Paramanand Singh Sep 9 '14 at 11:11
• Oh! I may be wrong, If possible I'll use a contradiction, If I'm wrong I apolize. – RE60K Sep 9 '14 at 11:12
• @Aditya: No need for apology, but I believe you are thinking in haste. Its basic algebra of limits $$\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$$ provided both limits on right side exist. Here in our question both the limits on right side are $0$ so that the limit of $f(x) - g(x)$ is also $0$. – Paramanand Singh Sep 9 '14 at 11:14
• Yes I was thinking of this, but a closer looks make me reveal that I'm wrong: $L_1=\frac{\sin x-x}{x^2}=-x/6+...\to0$ and $L_2=\frac{\cos x-1}x=-x/2+...$; I was thinking of $L_1+L_2/3$ which would cancel $x$ but that'll leave us with $x^2$ that also goes to zero.ii) As I was wrong I suddenly recongnized this basic algebra, thanks for reclarification – RE60K Sep 9 '14 at 11:20