Solution to Question 1:

We are tasked with proving that the sequence of functions $(f_n)_{n \in \mathbb{N}}$ defined as:

$f_n(x) = \sin(nx) \cdot e^{-n x^2} + \sqrt{1 - x^2}, \quad x \in [-1, 1], \; n \in \mathbb{N},$

converges pointwise on $[-1, 1]$ to a function $f(x)$.

Step 1: Analyze Each Term Separately

The function $f_n(x)$ has two terms: $f_n(x) = \underbrace{\sin(nx) \cdot e^{-n x^2}}_{\text{Term 1: Oscillatory}} + \underbrace{\sqrt{1 - x^2}}_{\text{Term 2: Constant}}.$

1. Second Term ($\sqrt{1 - x^2}$):

   • The term $\sqrt{1 - x^2}$ does not depend on $n$, so it remains constant as $n \to \infty$.
   • Therefore: $\lim_{n \to \infty} \sqrt{1 - x^2} = \sqrt{1 - x^2}, \quad \text{for all } x \in [-1, 1].$

2. First Term ($\sin(nx) \cdot e^{-n x^2}$):

   • The factor $e^{-n x^2}$ dominates the behavior of this term. For $x \neq 0$, observe:
     • $x^2 > 0$, so as $n \to \infty$, the exponential term $e^{-n x^2} \to 0$.
     • Since $|\sin(nx)| \leq 1$, the product $\sin(nx) \cdot e^{-n x^2} \to 0$ for all $x \neq 0$.
   • For $x = 0$, observe:
     • $e^{-n x^2} = e^{-n \cdot 0^2} = e^0 = 1$, so the term becomes $\sin(0) \cdot 1 = 0$.

Thus, for all $x \in [-1, 1]$, $\lim_{n \to \infty} \sin(nx) \cdot e^{-n x^2} = 0.$

Step 2: Combine the Results

Now, combining the limits of the two terms: $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \left[\sin(nx) \cdot e^{-n x^2} + \sqrt{1 - x^2}\right] = 0 + \sqrt{1 - x^2}.$

Thus, the sequence of functions $(f_n)$ converges pointwise to the function: $f(x) = \sqrt{1 - x^2}, \quad x \in [-1, 1].$

Conclusion

The sequence $(f_n)_{n \in \mathbb{N}}$ converges pointwise on $[-1, 1]$ to $f(x) = \sqrt{1 - x^2}$.

Would you like a detailed explanation of Questions 2 and 3?
Here are some related questions for further exploration:

1. How do we rigorously define pointwise convergence?
2. Why does the oscillatory term $\sin(nx)$ not affect the pointwise convergence?
3. What happens to $f_n(x)$ if we extend the domain outside $[-1, 1]$?
4. Can $\sqrt{1 - x^2}$ be approximated using Taylor series expansion near $x = 0$?
5. How does exponential decay in $e^{-n x^2}$ affect the convergence rate?

Tip: To understand pointwise vs. uniform convergence, compare how the speed of convergence depends on both $n$ and $x$.