Differentiation Question 508
Question: The value of $ \underset{n\to \infty }{\mathop{\lim }},[ \sqrt[3]{{{(n+1)}^{2}}}-\sqrt[3]{{{(n-1)}^{2}}} ] $ is
Options:
A) 1
B) -1
C) 0
D) $ -\infty $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{n\to \infty }{\mathop{\lim }},[ \sqrt[3]{{{(n+1)}^{2}}}-\sqrt[3]{{{(n-1)}^{2}}} ] $ $ =\underset{n\to \infty }{\mathop{\lim }},{n^{2/3}}[ {{( 1+\frac{1}{n} )}^{2/3}}-{{( 1-\frac{1}{n} )}^{2/3}} ] $ $ =\underset{n\to \infty }{\mathop{\lim }},{n^{2/3}}[ ( 1+\frac{2}{3}.\frac{1}{n}+\frac{\frac{2}{3}( \frac{2}{2}-1 )}{2!}\frac{1}{n^{2}}… )-( 1-\frac{2}{3}.\frac{1}{n}+\frac{\frac{2}{3}( \frac{2}{3}-1 )}{2!}\frac{1}{n^{2}}… ) ] $ $ =\underset{n\to \infty }{\mathop{\lim }},{n^{2/3}}[ \frac{4}{3}.\frac{1}{n}+\frac{8}{81}.\frac{1}{n^{3}}+… ] $ $ =[ \frac{4}{3}.\frac{1}{{n^{1/3}}}+\frac{8}{81}.\frac{1}{{n^{7/3}}}+… ]=0 $
BETA
