Question

If $f(x) = \sqrt{\frac{x - \sin x}{x + \cos^2 x}}$, then $\lim_{x\to\infty} f(x)$ is

• A. 0
• B. $\infty$
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

To find the limit of $f(x) = \sqrt{\frac{x - \sin x}{x + \cos^2 x}}$ as $x$ approaches infinity, we can analyze the expression inside the square root.

We have:

$\lim_{x\to\infty} \frac{x - \sin x}{x + \cos^2 x}$

Divide both the numerator and the denominator by $x$:

$\lim_{x\to\infty} \frac{1 - \frac{\sin x}{x}}{1 + \frac{\cos^2 x}{x}}$

Now, consider the limits of $\frac{\sin x}{x}$ and $\frac{\cos^2 x}{x}$ as $x$ approaches infinity.

Since $-1 \le \sin x \le 1$, we have $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$. As $x \to \infty$, $\frac{1}{x} \to 0$, so by the Squeeze Theorem, $\lim_{x\to\infty} \frac{\sin x}{x} = 0$.

Similarly, since $0 \le \cos^2 x \le 1$, we have $0 \le \frac{\cos^2 x}{x} \le \frac{1}{x}$. As $x \to \infty$, $\frac{1}{x} \to 0$, so by the Squeeze Theorem, $\lim_{x\to\infty} \frac{\cos^2 x}{x} = 0$.

Substituting these limits back into the expression:

$\lim_{x\to\infty} \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1$

Now, take the square root of the limit:

$\lim_{x\to\infty} f(x) = \sqrt{1} = 1$

Thus, the limit of $f(x)$ as $x$ approaches infinity is 1.