Question

$ \underset{x\to \infty }{\mathop{\lim }}\,\sqrt{\frac{x-\sin x}{x+{{\cos }^{2}}x}} $ is equal to

• A. $ -1 $
• B. $ 0 $
• C. $ \frac{1}{2} $
• D. $ 1 $

Answer 1

Answer: (D)

$ \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin x}{x}=0,\,\,\underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos x}{x}=0 $
Let $ A=\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{\frac{x-\sin x}{x+{{\cos }^{2}}x}} $
$ \Rightarrow $ $ A=\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{{{\cos }^{2}}x}{x}}} $
$ \Rightarrow $ $ A=\sqrt{\frac{1-0}{1+0}}=1 $