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Prove that the following limit does not exist.

$$

\lim_{x\to 0} \sin\left(1 \over x\right)

$$

Our definition of a limit:

- How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } $?
- $\lim_{n\rightarrow \infty}\int_0^1f_nhdm=\int_0^1fhdm$, prove $f\in L^p(m)$ , where $1\le p<\infty$.
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- Help with inf sup concept
- Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$
- Evaluating $\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$

Let $L$ be a number and let ${\rm f}\left(x\right)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for ever $\epsilon > 0$ there exists a corresponding $\delta > 0$ such that $0 < \left\vert\,x-c\,\right\vert \left\vert\,{\rm f}\left(x\right) – L\,\right\vert$

Not really sure how I go about doing this?

- Floor function as derivative
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- Approximating continuous functions with polynomials
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- $E \subset \mathbb R$ is an Interval $\iff E$ Is connected
- How to compute $\lim_{n\rightarrow\infty}\frac1n\left\{(2n+1)(2n+2)\cdots(2n+n)\right\}^{1/n}$
- Prove $\lim\limits_{n\to\infty}\frac{1}{\sqrt{n!}}=0$
- Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$
- How do I evaluate this integral?

**Hint**: Consider $x$ of the form

$$x_n = \frac{1}{2n\pi + \frac{\pi}{2}}$$

with $n$ a positive integer, and then consider

$$y_n = \frac{1}{2n\pi}$$

Then consider, say, $\epsilon = \frac{1}{2}$. Then for any $\delta$, there is an $n$ large enough that $x_n$ and $y_n$ are $\delta$-close to $0$.

You Can use sequential Criterion to show limit does not exists by showing two different sequence converging to $0$ but the functional sequences does not converges to the same limit

Suppose $k=\frac{1}{x}$. Your question would be the same as:

$\lim_{k\to\infty} \sin{(k)}$

Sine oscillates between -1 and 1, and does not have a limit

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