Constants a and b are chosen so that L=lim x→0(x+a xcos x-bsin x)/(x^5 exists and is finite. Find 240 L.

Expand the numerator with $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\dots$ and $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\dots$. It becomes $(1+a-b)x+\left(-\frac a2+\frac b6\right)x^3+\left(\frac a{24}-\frac b{120}\right)x^5+\dots$. For the limit to be finite the coefficients of $x$ and $x^3$ must vanish: $b=1+a$ and $b=3a$, so $a=\tfrac12$ and $b=\tfrac32$. Then $L=\frac1{48}-\frac1{80}=\frac{5-3}{240}=\frac1{120}$, so $240L=2$. The coefficients of $x$ and $x^3$ must vanish separately because the denominator is $x^5$, so any lower-order surviving term would make the limit infinite. Finally the $x^5$ coefficient is divided by $x^5$ and gives the limit, which we scale by 240 to obtain a single digit.