 Jett Castaneda

2022-01-26

Error evaluating $\underset{x\to 0}{lim}\frac{x-\mathrm{tan}x}{{x}^{3}}$
I solved it like this,
$\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{\mathrm{tan}x}{{x}^{3}}\right)=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{\mathrm{tan}x}{x}\cdot \frac{1}{{x}^{2}}\right)$
Now using the property
$\underset{x\to 0}{lim}\frac{\mathrm{tan}x}{x}=1$
we have:
$\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{1}{{x}^{2}}\right)=0$
Please explain my error! How can I avoid such errors? stamptsk

The explanation you're looking for is this. You are implicitly using the properties $lim\left(f+g\right)=lim\left(f\right)+lim\left(g\right)$ and $lim\left(fg\right)=lim\left(f\right)lim\left(g\right)$, but this is only true when all the limits in these equalities exist, (disclaimer: there are important assumptions that I'm not writing but that should accompany these properties). More specifically, what you did (implicitly) was:
$\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{\mathrm{tan}x}{{x}^{3}}\right)=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{\mathrm{tan}x}{x}\cdot \frac{1}{{x}^{2}}\right)$
$=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}\right)+\underset{x\to 0}{lim}\left(-\frac{\mathrm{tan}x}{x}\cdot \frac{1}{{x}^{2}}\right)$ (Incorrect)
$=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}\right)-\underset{x\to 0}{lim}\left(\frac{\mathrm{tan}x}{x}\right)\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}\right)$ (Incorrect)

$=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}\right)-\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}\right)\left(\ast \right)$
$=\underset{x\to 0}{lim}\left(\frac{1}{{x}^{2}}-\frac{1}{{x}^{2}}\right)$ (Incorrect)
$=0$ (**)
(*) As correct as something meaningless can be
(**) Actually correct, but it's too late Damian Roberts

Although, in fact, $\underset{x\to 0}{lim}\frac{\mathrm{tan}x}{x}=1$ you cannot deduce from that that
$\underset{x\to 0}{lim}\frac{1}{{x}^{2}}-\frac{\mathrm{tan}x}{x}×\frac{1}{{x}^{2}}=\underset{x\to 0}{lim}\frac{1}{{x}_{2}}-\frac{1}{{x}^{2}}$
In this case, LHopitals Rule is the way to go:
$\underset{x\to 0}{lim}\frac{x-\mathrm{tan}x}{{x}^{3}}=\underset{x\to 0}{lim}\frac{-{\mathrm{tan}}^{2}x}{3{x}^{2}}$
$=-{\frac{13}{\underset{x\to 0}{lim}\frac{\mathrm{tan}x}{x}}}^{2}$
$=-\frac{13}{}$

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