Mathematics - Limits Question with Solution | TestHub

Given ${\cos }^{-1}{\beta }_1+{\cos }^{-1}{\beta }_2+\dots +{\cos }^{-1}{\beta }_k=k\frac{\pi }{2}$. We know that ${\cos }^{-1}x\le \frac{\pi }{2}$ for all $x\ge -1$.

Therefore, ${\cos }^{-1}{\beta }_r\le \frac{\pi }{2}$ for all $r=1,2,3,\dots ,k$, which implies ${\sum }_{r=1}^k{\cos }^{-1}{\beta }_r\le \frac{k\pi }{2}$.

So, the given equality holds only if:

$${\cos }^{-1}{\beta }_1={\cos }^{-1}{\beta }_2=\dots ={\cos }^{-1}{\beta }_k=\frac{\pi }{2}$$

This implies ${\beta }_1={\beta }_2=\dots ={\beta }_k=0$.

Thus, $A={\sum }_{r=1}^k({\beta }_r{)}^r=0$.

The required limit is:

$$\underset{x\to 0}{\lim }\frac{(1+x^2{)}^{1/3}-(1-2x{)}^{1/4}}{x+x^2}$$

Applying L'Hôpital's Rule:

$$=\underset{x\to 0}{\lim }\frac{\frac{1}{3}(1+x^2{)}^{-2/3}(2x)-\frac{1}{4}(1-2x{)}^{-3/4}(-2)}{1+2x}$$

$$=\frac{0-\frac{1}{4}(-2)}{1+0}=\frac{\frac{1}{2}}{1}=\frac{1}{2}$$