Question

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{2 \mathrm{x}^{2 \mathrm{n}} \sin \frac{1}{\mathrm{x}}+\mathrm{x}}{1+\mathrm{x}^{2 \mathrm{n}}}$ then which of the following alternative(s) is/are correct ? (A) \(\lim _{x \rightarrow \infty} x f(x)=2\) (B) \(\lim \mathrm{f}(\mathrm{x})\) does not exist (C) \(\lim _{x \rightarrow 0} f(x)\) does not exist (D) \(\lim _{x \rightarrow-\gamma} \mathrm{f}(\mathrm{x})\) is equal to zero.

Short answer

Question: Determine the correct alternative(s) regarding the properties of the function \(f(x) = \lim_{n \rightarrow \infty} \frac{2x^{2n} \sin{\frac{1}{x}}+x}{1+x^{2n}}\). Answer: (A) and (D)

Step-by-step solution

Case 1: x -> 0

As x approaches 0, x^(2n) will also approach 0 since exponent n is approaching infinity. So, inside the limit, the function becomes: \(\lim_{n \rightarrow \infty} \frac{2x^{2n} \sin{\frac{1}{x}}+x}{1+x^{2n}} = \lim_{n \rightarrow \infty} \frac{x}{1+x^{2n}}\) Now, applying limit x -> 0 on this simplified expression: \(\lim_{x \rightarrow 0} \lim_{n \rightarrow \infty} \frac{x}{1+x^{2n}} = 0\) So, the limit at 0 exists.

Case 2: x -> infinity

As x approaches infinity, x^(2n) will also approach infinity. So, inside the function, we have: \(\lim_{n \rightarrow \infty} \frac{2x^{2n} \sin{\frac{1}{x}}+x}{1+x^{2n}}\) Dividing by x: \(\lim_{n \rightarrow \infty} \frac{2x^{2n-1} \sin{\frac{1}{x}}+1}{\frac{1}{x}+x^{2n-1}}\) Now, applying limit x -> infinity on this simplified expression: \(\lim_{x \rightarrow \infty} \lim_{n \rightarrow \infty} \frac{2x^{2n-1} \sin{\frac{1}{x}}+1}{\frac{1}{x}+x^{2n-1}} = 2\) So, the limit at infinity exists and is equal to 2.

Case 3: x -> -infinity

As x approaches -infinity, x^(2n) will still approach infinity (since even n's raise negative values to positive). So, inside the function, we have: \(\lim_{n \rightarrow \infty} \frac{2x^{2n} \sin{\frac{1}{x}}+x}{1+x^{2n}}\) Applying limit x -> -infinity (the behavior is the same as Case 2 because x^(2n) becomes positive): \(\lim_{x \rightarrow -\infty} \lim_{n \rightarrow \infty} \frac{2x^{2n} \sin{\frac{1}{x}}+x}{1+x^{2n}} = \lim_{x \rightarrow -\infty} \frac{x}{1+x^{2n}} = 0\) So, the limit at -infinity exists and is equal to 0. Now, comparing our results with given options: (A) \(\lim _{x \rightarrow \infty} xf(x) = 2\) - True (B) \(\lim f(x)\) does not exist - False (C) \(\lim _{x \rightarrow 0} f(x)\) does not exist - False (D) \(\lim _{x \rightarrow-\infty} f(x)\) is equal to zero - True Hence, the correct alternative(s) are (A) and (D).