Question

Let $0\le {\beta }_r\le 1$ and $\sum _{r=1}^k{\cos }^{-1}{\beta }_r=\frac{k\pi }{2}$ for any $k\ge 1$ and $A=\sum _{r=1}^k({\beta }_r{)}^r$, then
$\underset{x\to A}{lim}498.\frac{(1+x^2{)}^{1/3}-(1-2x{)}^{1/4}}{x+x^2}$ is equal to

Answer 1

Since maximum value of ${\cos }^{-1}x$ in $[0,1]=\pi /2,\sum _{r=1}^k{\cos }^{-1}{\beta }_r=k\pi /2$ is possible if and only if each ${\cos }^{-1}{\beta }_r=\pi /2\iff {\beta }_r=0$
$A=\sum _{r=1}^k({\beta }_r{)}^r=0$
Thus $\underset{x\to 0}{lim}\frac{(1+x^2{)}^{1/3}-(1-2x{)}^{1/4}}{x+x^2}$
$=\underset{x\to 0}{lim}\frac{(1+x^2/3+0(x^4\left)\right)-(1-x/2+0(x^3\left)\right)}{x(1+x)}$
$=\underset{x\to 0}{lim}\frac{1/2+x/3+0\left(x^2\right)}{1+x}=1/2$. Thus the reqd. limit is 249.