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Recently, I have been struggling with this one problem: $$\lim _{x\rightarrow 0}\left(\frac{1}{\sin( x)} -\frac{1}{x}\right)$$ and cannot figure out a way to solve it without the use of L'Hôpital's rule.

The only thing I can think of using is the basic identity $$\lim_{x\to 0}\left(\frac{\sin( x)}{x}\right) =1$$ but I can't reduce the original problem down to a point where I can apply this identity.

Any help would be greatly appreciated.

Thanks!

Blue
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SacredCobalt
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  • Just use $\frac{1}{\sin x} -\frac{1}{x} = \frac{1}{x} \cdot \frac{x}{\sin x} - \frac{1}{x}$ –  Sep 19 '20 at 01:05
  • https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Sep 19 '20 at 01:18

3 Answers3

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\begin{align} \lim _{x\rightarrow 0}\left(\frac{1}{\sin( x)} -\frac{1}{x}\right)&=\lim _{x\rightarrow 0}\left(\frac{x-\sin(x)}{x^2}\cdot\frac{x}{\sin(x)}\right)\\ &=\lim _{x\rightarrow 0}\left(\frac{x-\sin(x)}{x^2}\right)\\ \end{align} Now, use the Taylor series of $\sin(x)$.

Kenta S
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Without L'Hopital

We can add the two fractions together.

$\frac {x - \sin x}{x\sin x}$

And now we need to get a little bit creative.

$|\sin x| \le |x| \le |\tan x|$ (and all have the same sign when $-\frac {\pi}{2} <x< \frac {\pi}{2})$

$|x - \sin x| \le |\tan x - \sin x|$

With this substitution we can find some limits we know how to resolve.

$|\lim_\limits{x\to 0} \frac {x - \sin x}{x\sin x}| \le |\lim_\limits{x\to 0} \frac {\tan x - \sin x}{x\sin x}|$

$\lim_\limits{x\to 0} \frac {\tan x - \sin x}{x\sin x}\\ \lim_\limits{x\to 0} \frac {\sec x - 1}{x}\\ \lim_\limits{x\to 0} \left(\frac {1 - \cos x}{x}\right) \sec x = 0$

And by the squeeze theorem. $0\le |\lim_\limits{x\to 0} \frac {1}{\sin x} - \frac {1}{x}| \le \sec x \lim_\limits{x\to 0} \frac {1 - \cos x}{x}$

Gives us $0.$

Doug M
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You could use the first few terms of the Taylor series for $\sin(x)$ around $x = 0$ (i.e., the Maclaurin series) in the numerator to get

$$\begin{equation}\begin{aligned} \lim_{x \rightarrow 0}\left(\frac{1}{\sin( x)} - \frac{1}{x}\right) & = \lim_{x \rightarrow 0}\left(\frac{x - \sin(x)}{x\sin(x)}\right) \\ & = \lim_{x \rightarrow 0}\left(\frac{x - (x - \frac{x^3}{3} + O(x^5))}{x\sin(x)}\right) \\ & = \lim_{x \rightarrow 0}\left(\frac{\frac{x^2}{3} + O(x^4))}{\sin(x)}\right) \\ & = \lim_{x \rightarrow 0}\left(\frac{x}{\sin(x)}\right)\left(\frac{x}{3} + O(x^3))\right) \\ & = 1(0) \\ & = 0 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

John Omielan
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