Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
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: Evaluate $\lim_{n \to \infty} \frac{2x^{2n} \sin\frac{1}{x} + x}{1 + x^{2n}}$ by analyzing behavior of $x^{2n}$ for different ranges: when $|x| > 1$, $x^{2n} \to \infty$; when $|x| < 1$, $x^{2n} \to 0$; when $|x| = 1$, special handling required.
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