Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
Let $f(x) = \lim_{n \to \infty} \frac{2x^{2n} \sin + x}{1 + x^{2n}}$ then which of the following alternative(s) is/are correct?
lim_{x \to \infty} f(x) = 2
lim_{x \to 1} f(x) does not exist
lim_{x \to 0} f(x) does not exist
lim_{x \to \infty} f(x) is equal to zero
Step-by-Step Solution
Key Concept: When limits depend on the growth rate of exponential terms ($x^{2n}$), split into cases based on whether $|x|$ is greater than, equal to, or less than 1.
Using the limit $\lim_{n \to \infty} \frac{2y^{2n}\sin\frac{1}{x} + x}{1 + x^{2n}}$, we analyze by cases. For $|x| > 1$ or $x < -1$: dividing numerator and denominator by $x^{2n}$ gives $f(x) = 2\sin\frac{1}{x}$. At $x = 1$: the limit equals $\frac{2(\sin 1) - 1}{2}$. At $x = -1$: the limit equals $\frac{-2(\sin 1) - 1}{2}$. For $|x| < 1$: the denominator dominates and $f(x) = x$. We verify: $\lim_{x \to \infty} xf(x) = 2$, $\lim_{x \to 1^-}f(x) = 2\sin 1 \ne 1$, so the limit at $x=1$ does not exist. However $\lim_{x \to 0} f(x) = 0$ exists.
Correct Answer: 1,2,4