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 \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}=\dfrac{0}{\begin{align}
  & 0 \\
 & \\
\end{align}}\] 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.