Question

If $\sum\limits_{r=1}^{k}{{{\cos }^{-1}}{{\beta }_{r}}=\dfrac{k\pi }{2}}$ for any $k\ge 1$ and $A=\sum\limits_{r=1}^{k}{{{\left( {{\beta }_{r}} \right)}^{r}}}$, then $\underset{x\to A}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}$ is equal to
(a). $0$
(b). $\dfrac{1}{2}$
(c). $\dfrac{\pi }{2}$
(d). $\dfrac{5}{6}$

Answer 1

Hint: Use the fact that the value of the function ${{\cos }^{-1}}x$ is $\dfrac{\pi }{2}$ for $x=0$. Then find the value of ${{\beta }_{r}}$ and thus $A$. Finally evaluate the limit using the L'Hospital Rule.

Complete step-by-step answer:
We have the function $\sum\limits_{r=1}^{k}{{{\cos }^{-1}}{{\beta }_{r}}=\dfrac{k\pi }{2}}$. We need to find the value of ${{\beta }_{r}}$. We observe that the sum of $k$ terms is $\dfrac{k\pi }{2}$. Thus, the value of ${{\cos }^{-1}}{{\beta }_{r}}$ is $\dfrac{\pi }{2}$ for each $r$. Hence, for each $r$, we have ${{\beta }_{r}}=0$.
Now, we will use this to evaluate the value of the function $A=\sum\limits_{r=1}^{k}{{{\left( {{\beta }_{r}} \right)}^{r}}}$.
Thus, we have $A=\sum\limits_{r=1}^{k}{{{\left( {{\beta }_{r}} \right)}^{r}}}={{0}^{1}}+{{0}^{2}}+{{0}^{3}}+{{...0}^{k}}=k\left( 0 \right)=0$.
Hence, we have $A=0$.
Now, we will evaluate the limit $\underset{x\to A}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}$. As $A=0$, we have $\underset{x\to A}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}$.
We observe that if we apply the limit directly, we get $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}=\dfrac{1-1}{0+0}=\dfrac{0}{0}$.
Hence, we will use L’Hopital Rule to evaluate the given limit of the function which states that if $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}=\dfrac{0}{\begin{aligned} & 0 \\\ & \\\ \end{aligned}}$ then we will use $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}$ to evaluate the limit.
Substituting $f\left( x \right)={{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}},g\left( x \right)=x+{{x}^{2}}$ in the above equation, we have $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)}{\dfrac{d}{dx}(x+{{x}^{2}})}$.
We can rewrite the above equation as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)}{\dfrac{d}{dx}(x+{{x}^{2}})}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}} \right)-\dfrac{d}{dx}\left( {{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)}{\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)}.....\left( 1 \right)$. We know that the derivative of the function of the form $y={{(ax+b)}^{n}}$ is $\dfrac{dy}{dx}=na{{\left( ax+b \right)}^{n-1}}$.
Substituting $a=1,b=1,n=\dfrac{1}{3}$ in the above formula, we have $\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}} \right)=\dfrac{1}{3}{{\left( x+1 \right)}^{\dfrac{-2}{3}}}.....\left( 2 \right)$.
Substituting $a=-2,b=1,n=\dfrac{1}{4}$ in the above formula, we have $\dfrac{d}{dx}\left( {{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)=\dfrac{-1}{2}{{\left( 1-2x \right)}^{\dfrac{-3}{4}}}.....\left( 3 \right)$.
Substituting $a=1,b=0,n=1$ in the above formula, we have $\dfrac{d}{dx}\left( x \right)=1.....\left( 4 \right)$.
Substituting $a=1,b=0,n=2$ in the above formula, we have $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x.....\left( 5 \right)$.
Substituting equation $\left( 2 \right),\left( 3 \right),\left( 4 \right)$and$\left( 5 \right)$ in equation $\left( 1 \right)$, we get $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)}{\dfrac{d}{dx}(x+{{x}^{2}})}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\left( 1+x \right)}^{\dfrac{1}{3}}} \right)-\dfrac{d}{dx}\left( {{\left( 1-2x \right)}^{\dfrac{1}{4}}} \right)}{\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{3}{{\left( 1+x \right)}^{\dfrac{-2}{3}}}-\left( \dfrac{-1}{2} \right){{\left( 1-2x \right)}^{\dfrac{-3}{4}}}}{1+2x}$
On further solving the equation by applying limits, we have $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{3}{{\left( 1+x \right)}^{\dfrac{-2}{3}}}+\dfrac{1}{2}{{\left( 1-2x \right)}^{\dfrac{-3}{4}}}}{1+2x}=\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1+0}=\dfrac{5}{6}$.
Hence, the value of $\underset{x\to A}{\mathop{\lim }}\,\dfrac{{{\left( 1+x \right)}^{\dfrac{1}{3}}}-{{\left( 1-2x \right)}^{\dfrac{1}{4}}}}{x+{{x}^{2}}}$is$\dfrac{5}{6}$.

Note: We can take other possible values of ${{\beta }_{r}}$ as well, however, for each value that satisfies the given condition, we will get $A=0$ and hence, the limit will be the same.