Finding $\lim\limits_{n→∞}\frac{n^p \sin^2(n!)}{n+1}$

Question

It is given that $p > 0$. I did this question by first taking $n$ common and then I was getting answer as infinity but the answer in the book is given as $0$.

Answer 1

Let's start with $\sin^2(n!)$. It will fluctuate in the interval $(0, 1)$ (open interval). It will never be 0 or 1 for any $n > 0$ because $n! \neq k \pi/2$ for any $n$ and any integer $k$. Anyway, it is bounded and positive, and it will only have an impact on the limit if the rest of the expression converges to neither 0 nor $\pm \infty$.

We can rewrite the rest, $$\frac{n^p}{n+1} = \frac{n^{p-1}}{1+1/n},$$ where we see that the denominator will approach 1 as $n$ approaches $\infty$. Hence, to understand the limit we now only need to understand the behaviour of $n^{p-1}$. For $0 < p < 1$, it's limit is 0 and hence the limit of the entire expression is also 0. For $p = 1$ the limit is undefined because $n^0 = 1$ and the only factor not equal to 1 is $\sin^2(n!)$, which fluctuates. For $p > 1$, the limit is $\infty$.

Your textbook is correct for $0 < p < 1$, but not for $p \ge 1$.