Functions Question 162
Question: The value of $ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{2}\sin \frac{1}{x}-x}{1-|x|} $ is
Options:
A) $ 0 $
B) 1
C) ?1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Putting $ x=\frac{1}{t}, $ the given limit $ =\underset{t\to 0}{\mathop{\lim }},\frac{\frac{\sin t}{t}-1}{t-1}=\frac{1-1}{0-1}=0, $ which is given in . Aliter : $ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{2}\sin \frac{1}{x}-x}{1-|x|} $ $ =\underset{x\to \infty }{\mathop{\lim }},\frac{x^{2},( \frac{1}{x}-\frac{1}{3!}\frac{1}{x^{3}}+…. )-x}{1-|x|} $ , $ [ \because \frac{1}{x}\to 0 ] $ $ =\underset{x\to \infty }{\mathop{\lim }},\frac{( x-\frac{1}{6x}+….-x )}{1-|x|} $ $ =\underset{x\to \infty }{\mathop{\lim }}\frac{\frac{1}{6x}-\text{terms containing powers of }\frac{1}{x}}{|x|-1}=0. $
BETA
