Question

Let $\lim_{n\to\infty}\frac{1+cn^2}{(2n+3+2\sin n)^2}=\frac{1}{2}$. If $c\le\alpha\le\beta$ where $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-2px+p^2-1=0$, then find the minimum integral value of p.

Answer 1

3

Answer 2

The given limit is $\lim_{n\to\infty}\frac{1+cn^2}{(2n+3+2\sin n)^2}$.

To evaluate this limit, we divide the numerator and the denominator by the highest power of $n$ in the denominator, which is $n^2$.

The expression becomes $\frac{\frac{1}{n^2}+c}{\left(\frac{2n+3+2\sin n}{n}\right)^2} = \frac{\frac{1}{n^2}+c}{\left(2+\frac{3}{n}+\frac{2\sin n}{n}\right)^2}$.

Taking the limit as $n\to\infty$:

$\lim_{n\to\infty} \frac{1}{n^2} = 0$

$\lim_{n\to\infty} \frac{3}{n} = 0$

$\lim_{n\to\infty} \frac{2\sin n}{n} = 0$ (since $-1 \le \sin n \le 1$, so $-\frac{2}{n} \le \frac{2\sin n}{n} \le \frac{2}{n}$. By the Squeeze Theorem, as $n\to\infty$, $\frac{2\sin n}{n} \to 0$).

So, the limit is $\frac{0+c}{(2+0+0)^2} = \frac{c}{4}$.

We are given that the limit is equal to $\frac{1}{2}$.

$\frac{c}{4} = \frac{1}{2} \implies c = 4 \times \frac{1}{2} = 2$.

The quadratic equation is $x^2-2px+p^2-1=0$.

We can rewrite this as $x^2-2px+p^2 = 1$, which is $(x-p)^2 = 1$.

Taking the square root of both sides, $x-p = \pm 1$.

The roots are $x = p \pm 1$.

Let $\alpha = p-1$ and $\beta = p+1$. Since $p-1 \le p+1$, this assignment satisfies $\alpha \le \beta$.

We are given the condition $c \le \alpha \le \beta$.

Substituting $c=2$, $\alpha=p-1$, and $\beta=p+1$, we get:

$2 \le p-1 \le p+1$.

This inequality can be split into two parts:

1. $2 \le p-1$

Adding 1 to both sides gives $3 \le p$, or $p \ge 3$.

2. $p-1 \le p+1$

Subtracting $p$ from both sides gives $-1 \le 1$, which is always true and gives no additional constraint on $p$.

So, the condition $c \le \alpha \le \beta$ is equivalent to $p \ge 3$.

We are asked to find the minimum integral value of $p$.

The integers that satisfy $p \ge 3$ are $3, 4, 5, \dots$.

The minimum integral value among these is 3.