Jessie Lee

2021-12-30

Evaluating $\underset{x\to 0}{lim}\left(\frac{1}{\mathrm{sin}\left(x\right)}-\frac{1}{x}\right)$ without L'Hôpital's rule
The only thing I can think of using is the basic identity
$\underset{x\to 0}{lim}\left(\frac{\mathrm{sin}x}{x}\right)=1$
but I can't reduce the original problem down to a point where I can apply this identity.

deginasiba

$\underset{x\to 0}{lim}\left(\frac{1}{\mathrm{sin}x}-\frac{1}{x}\right)=\underset{x\to 0}{lim}\left(\frac{x-\mathrm{sin}x}{{x}^{2}}\cdot \frac{x}{\mathrm{sin}x}\right)$
$=\underset{x\to 0}{lim}\left(\frac{x-\mathrm{sin}x}{{x}^{2}}\right)$
Now, use the Taylor series of $\mathrm{sin}x$

Medicim6

Without LHopital

Vasquez

You could use the first few terms of the Taylor series for $\mathrm{sin}\left(x\right)$ around x=0 (i.e., the Maclaurin series) in the numerator to get
$\begin{array}{}\underset{x\to 0}{lim}\left(\frac{1}{\mathrm{sin}x}-\frac{1}{x}\right)=\underset{x\to 0}{lim}\left(\frac{x-\mathrm{sin}x}{x\mathrm{sin}x}\right)\\ =\underset{x\to 0}{lim}\left(\frac{x-\left(x-\frac{{x}^{3}}{3}+O\left({x}^{5}\right)\right)}{x\mathrm{sin}x}\right)\\ =\underset{x\to 0}{lim}\left(\frac{\frac{{x}^{2}}{3}+O\left({x}^{4}\right)}{\mathrm{sin}x}\right)\\ =\underset{x\to 0}{lim}\left(\frac{x}{\mathrm{sin}x}\right)\left(\frac{x}{3}+O\left({x}^{3}\right)\right)\\ =1\left(0\right)\\ =0\end{array}$

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