Question

\(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}+x^{2} \sin (1 / x)}\) is equal to (A) 0 (B) 2 (C) \(-2\) (D) none of these

Short answer

Answer: The limit of the expression as x approaches negative infinity is \(-2\).

Step-by-step solution

Identify the dominating term

As x approaches negative infinity, we can observe that the dominating term in the expression is \(x^2\). We can then focus on finding the limit with respect to this term.

Factor out \(x^2\) from the expression

Let's factor out \(x^2\) from the expression inside the square root: \(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}(1+\sin (1 / x ))}\)

Simplify the expression further

Now, we can rewrite the above expression into a simpler form: \(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}}\sqrt{(1+\sin (1 / x ))}\)

Analyze the behavior of \(\sin(1/x)\) as x approaches negative infinity

As x approaches negative infinity, \(1/x\) approaches 0, which means the argument of the sine function oscillates between close to -1 and close to 1. Thus, we can write the inequality: -1 ≤ \(\sin (1 / x)\) ≤ 1 which gives us the inequality for the expression inside the square root: 0 ≤ \(1+\sin (1 / x)\) ≤ 2

Apply the square root to the inequality

Since the square root function is monotonic, we can take the square root of each term of the inequality: \(\sqrt{0}\) ≤ \(\sqrt{1+\sin (1 / x)}\) ≤ \(\sqrt{2}\)

Find the limit of the expression

Now we can take the limit of each term of the inequality: \(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}}\sqrt{0}\) ≤ \(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}}\sqrt{(1+\sin (1 / x ))}\) ≤ \(\lim _{x \rightarrow-\infty} x+\sqrt{x^{2}}\sqrt{2}\)

Simplify and evaluate the limit

As x approaches negative infinity, the limit of each term yields the following: \(-2\) ≤ \(-\infty + \sqrt{x^2}\sqrt{(1+\sin (1 / x ))}\) ≤ \(-\infty + \sqrt{2x^2}\) Since the middle term is trapped between -2 and -∞ as x approaches negative infinity, we can conclude that the limit is equal to -2. Hence, the correct answer is (C) \(-2\).