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Is it valid to say the following?

$$\lim_{x \to 0} \frac{0}{x}=0$$

It seems like it is since $\frac{0}{x} \leq \frac{x^2}{x}$ which clearly converges to $0$. However, I thought $\frac{0}{0}$ was undefined?

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    $\begingroup$ Sorry, I'm in high school and self taught so I'm not sure what you mean by that. $\endgroup$ – smurfette25 Apr 17 '14 at 9:20
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    $\begingroup$ Suppose $\epsilon > 0$. You want to find $\delta > 0$ such that if $|x| < \delta$, then $| \frac{0}{x} | < \epsilon $. But, $| \frac{0}{x} | = 0 < \epsilon$. So, no matter what $\delta $ you choose, you will obtain $0$ as the limit. $\endgroup$ – user139708 Apr 17 '14 at 9:22
  • $\begingroup$ Isn't that the definition of continuity though? Can you explain the connection between being continuous and the limit? Thanks! $\endgroup$ – smurfette25 Apr 17 '14 at 9:24
  • $\begingroup$ $\mathbf{Definition}$: We say $$ \lim_{x \to a} f(x) = L$$ if for any given $\epsilon > 0$, there exists $\delta > 0$ such that if $|x-a| < \delta$, then $|f(x)-L|<\epsilon $ $\endgroup$ – user139708 Apr 17 '14 at 9:24
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    $\begingroup$ The function you've chosen, $0/x$, is undefined at the origin, and zero everywhere else. So the limit at the origin must equal zero (obviously this is an informal argument, but that's the intuition for limits). $\endgroup$ – Spine Feast Apr 17 '14 at 9:31
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Hint: For every $x$ in a punctured neighborhood of $0$ the expression $\frac{0}{x}$ equals zero exactly.

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  • $\begingroup$ So, I can say for $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n > N$, $d(x,0) < \epsilon$? $\endgroup$ – smurfette25 Apr 17 '14 at 9:22
  • $\begingroup$ @bobb42 Not quite, it's $d(0/x,0)<\epsilon$. $\endgroup$ – user1337 Apr 17 '14 at 9:26
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Suppose $ϵ>0$. You want to find δ>0 such that if $0<|x|<δ$, then $|\frac{0}{x}|<ϵ$. But,$ \frac{0}{x} =0<ϵ$. So, no matter what $δ$ you choose, you will obtain $0$ as the limit.

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