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Chapter 1: Problem 111

Let $\mathrm{f}(\mathrm{x})=\left\\{\begin{array}{ll}{\left[\mathrm{x}^{\mathrm{x}}\right]+\mathrm{a},} & \mathrm{x}>0 \\ \lim _{t \rightarrow \infty}\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)^{\mathrm{t}}, & \mathrm{x}<0\end{array}\right.$. The complete set of the values of ' \(\mathrm{a}\) ' for which Approx exists is (A) \((0,2]\) (B) \((-2,2)\) (C) \([-1,1]\) (D) None of these

Short Answer

Expert verified
Answer: (D) None of these. The only possible value for 'a' is 0.

Step by step solution

01

Determine the limit of f(x) as x approaches 0 from the right side (x > 0).

For x > 0, the function is given as f(x) = [x^x] + a. The limit as x approaches 0 from the positive side can be found using L'Hopital's rule and then adding the value of 'a': $$ \lim_{x \to 0^+} ([x^x] + a) = \lim_{x \to 0^+} (0 + a) = a $$
02

Determine the limit of f(x) as x approaches 0 from the left side (x < 0).

For x < 0, the function is given as f(x) = lim_{t \to \infty} \left(\frac{\sin x}{x}\right)^t. To find the limit as x approaches 0 from the left side, we can use L'Hopital's rule: $$ \lim_{x \to 0^-} \lim_{t \to \infty} (\frac{\sin x}{x})^t = \lim_{t \to \infty}(\frac{\sin 0^-}{0})^t = \lim_{t \to \infty}(0)^t = 0 $$
03

Set the left-hand and right-hand limits equal.

To ensure continuity at x = 0, we should have the left-hand limit equal to the right-hand limit: $$ a = 0 $$
04

Find the possible values of 'a'.

Since the only equation we got is a = 0, it means there is only one possible value for 'a' to ensure continuity, which is a = 0. This does not match any option given: - (A) \((0,2]\) - (B) \((-2,2)\) - (C) \([-1,1]\) So, the correct answer is: (D) None of these.

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Most popular questions from this chapter

If $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right)$ $=\mathrm{K} \cdot \lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})\( then the value of \)\mathrm{K}$ (A) (B) (C) \(2 \mathrm{a}\) (D) None of these

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