# \lim_{x \to 0} \frac{x(1+a \cos x)-b \sin x}{x^3}=1, how to

$\underset{x\to 0}{lim}\frac{x\left(1+a\mathrm{cos}x\right)-b\mathrm{sin}x}{{x}^{3}}=1$, how to find the constants a,b?
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Vivian Soares
If you can use Taylor series, the numerator is
$x\left(1+a-\frac{a{x}^{2}}{2}\right)-b\left(x-\frac{{x}^{3}}{6}\right)+o\left({x}^{3}\right)=\left(1+a-b\right)x+\left(b-3a\right)\frac{{x}^{3}}{6}+o\left({x}^{3}\right)$
Solving $1+a-b=0,b-3a=6$ gives $a=-\frac{52}{,}b=-\frac{32}{}$
###### Not exactly what you’re looking for?
kaluitagf
Apply LHopitals rule again:
$\underset{x\to 0}{lim}\frac{x\left(1+a\mathrm{cos}x\right)-b\mathrm{sin}x}{{x}^{3}}\stackrel{LHR}{=}\underset{x\to 0}{lim}\frac{1+\left(a-b\right)\mathrm{cos}x-ax\mathrm{sin}x}{3{x}^{2}}$
$\stackrel{LHR}{=}\underset{x\to 0}{lim}\frac{\left(b-2a\right)\mathrm{sin}x-ax\mathrm{cos}x}{6x}$
and recall that $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$