Karina Trujillo

2022-06-26

For what values of a and b is the following equation true?
$\underset{x\to 0}{lim}\left(\frac{\mathrm{sin}\left(2x\right)}{{x}^{3}}+a+\frac{b}{{x}^{2}}\right)=0$

Expert

We have
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(2x\right)}{{x}^{3}}+a+\frac{b}{{x}^{2}}\\ & =\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(2x\right)+a{x}^{3}+bx}{{x}^{3}}\end{array}$
which is a limit of type $0/0$, so we try L'Hôpital's rule:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{2\mathrm{cos}\left(2x\right)+3a{x}^{2}+b}{3{x}^{2}}\end{array}$
The denominator again vanishes for $x\to 0$ , the nominator goes to $2+b$. So if $b\ne -2$, the nominator does not vanish and we have
$L=\mathrm{sgn}\left(2+b\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{\infty }$
For $b=-2$ we again have a limit of type $0/0$ and apply the rule again:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{-4\mathrm{sin}\left(2x\right)+6ax}{6x}\end{array}$
The nominator and denominator vanish and we apply the rule once again:
$\begin{array}{rl}L& =\underset{x\to 0}{lim}\frac{-8\mathrm{cos}\left(2x\right)+6a}{6}=\frac{6a-8}{6}=a-\frac{4}{3}\end{array}$
This gives the answer that the equation is true, L vanishes, if $a=4/3$ and $b=-2$

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