Question-32
\lim _{x \rightarrow \infty} \sqrt[3]{x}\left(\sqrt[3]{(x+1)^{2}}-\sqrt[3]{(x-1)^{2}}\right)=A) 1/3
B) 2/3
C) 1
D) 4/3
Answer: D) 4/3
Explanation
L=\lim _{x \rightarrow \infty} \sqrt[3]{x}\left(\sqrt[3]{(x+1)^{2}}-\sqrt[3]{(x-1)^{2}}\right)We know that:
\begin{aligned} & a^{2}-b^{2}=(a-b)(a+b) \\ & \therefore L=\lim _{x \rightarrow \infty} x^{\frac{1}{3}}\left\{(x+1)^{\frac{1}{3}}+(x-1)^{\frac{1}{3}}\right\}\left\{(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}\right\} \end{aligned}Multiplying and dividing by:
\left\{(x+1)^{\frac{2}{3}}+(x+1)^{\frac{1}{3}}(x-1)^{\frac{1}{3}}+(x-1)^{\frac{2}{3}}\right\}Use:
a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right) \left((x+1)^\frac13\right)^3-\left((x-1)^\frac13\right)^3 =\left\{(x+1)^\frac13-(x-1)^\frac13\right\}\left\{(x+1)^\frac23+(x+1)^\frac13(x-1)^\frac13+(x-1)^\frac23\right\} \therefore L=\lim_{x\rightarrow\infty}\frac{x^\frac13\left\{(x+1)^\frac13+(x-1)^\frac13\right\}^2}{\left\{(x+1)^\frac23+\left(x^2-1\right)^\frac13+(x-1)^\frac23\right\}}Taking common x from all the brackets, we get:
or, L=\lim _{x \rightarrow \infty} \frac{2 x^{\frac{2}{3}}\left\{\left(1+\frac{1}{x}\right)^{\frac{1}{3}}+\left(1-\frac{1}{x}\right)^{\frac{1}{3}}\right\}}{x^{\frac{2}{3}}\left\{\left(1+\frac{1}{x}\right)^{\frac{2}{3}}+\left(1-\frac{1}{x^{2}}\right)^{\frac{1}{3}}+\left(1-\frac{1}{x}\right)^{\frac{2}{3}}\right\}}
Now, put limit x \rightarrow \infty, so we get:
L=\frac{4}{3}