Find $$ \lim_{n \to \infty}\left\{2n^2\cos\left(1 \over n\right)\right\} \cos\left(1 \over n\right) $$ where $\left\{x\right\}= x - \left\lfloor x\right\rfloor$ is the fractional part.
Please help me solve it. I know that $\cos\left(n\right) < n$ and that this applies also for $\sin$.
On
This is fiendish.
First, of course, the factor $\cos\frac1n$ changes nothing, since its limit as $n\to\infty$ is $1$. Thus we concentrate on the other factor, involving Fractional Part.
Now, consider an integer that’s large, $n$. Then $$\cos\frac1n=1-\frac1{2n^2}+\frac1{24n^4}-\star\quad, $$ where the “$\star$” represents something really tiny, and ignorable in comparison with the $\frac1{24n^4}$-term. Now, when we multiply the displayed quantity by $2n^2$, what’s its fractional part? It’s just the $\frac1{24n^4}-\star$, which is positive and small, approaching zero as $n\to\infty$. Of course this calculation depends strongly on $n$ being an integer.
Thus the limit exists, and is zero.
Let $f(x)=2x^2\cos(\frac{1}{x})$ for $x>0$.
It is easy to see that $f$ is continous in $\mathbb{R}^{+}$.
Since, $\lim_{n\rightarrow \infty}f(n) = \infty$, we have, by intermediate value theorem, $\lim_{n\rightarrow\infty}\{f(n)\}$ does not exist.
Therefore, $\lim_{n\rightarrow\infty}\{f(n)\}.\cos(\frac{1}{n})$ does not exist. As, $\lim_{n\rightarrow\infty}\cos(\frac{1}{n}) = 1$
Justification of above argument:
If $\lim_{n\rightarrow\infty}\{f(n)\}.\cos(\frac{1}{n})$ exists, let $P(n) = \{f(n)\}.\cos(\frac{1}{n})$.
Then, $\lim_{n\rightarrow\infty}\frac{P(n)}{\cos(\frac{1}{n})} = \lim_{n\rightarrow\infty}P(n)$. As, $\lim_{n\rightarrow\infty}\cos(\frac{1}{n}) = 1 \neq 0$