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I need help to show the continuity of $f(x)=\frac{\sin(x)}{x}$ (for every $x$ except $0$, let $f(0)$ be $1$) Thanks!

3 Answers 3

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This reduces down to an old calculus fact.

Since you know, by facts about dividing continuous functions, that $\displaystyle \frac{\sin(x)}{x}$ is continuous everywhere it's defined, i.e. everywhere but $0$. So, to check continuity of $f$ you really only need to check at $0$. This boils down to verify that $\displaystyle 1=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\sin(x)}{x}$.

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    Okay, got your point. But now, how to verify that $\displaystyle 1=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\sin(x)}{x}$?2012-12-09
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Showing that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ can be done in a lot of different ways. Here's one such proof that doesn't rely explicitly on L'hopital. Since $\sin(x)$ is differentiable at $0$, we can write that $\sin(x) = x + \epsilon(x)$ where $\frac{\epsilon(x)}{x} \to 0$ as $x \to 0$. So $\frac{\sin(x)}{x} = \frac{x + \epsilon(x)}{x} = 1 + \frac{\epsilon(x)}{x} \to 1$, as $x \to 0$.

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    Not if you define it in terms of power series, or $e^x$. How else would you define $\sin(x)$?2012-12-09
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An other way to show that $\lim_{x → 0} \frac{\sin(x)}{x} = 1$.

You probably learned that $f'(a)=\lim_{x → a} \frac{f(x)-f(a)}{x-a}$. Therefore:

$\lim_{x → 0} \frac{\sin(x)}{x}=\lim_{x → 0} \frac{\sin(x)-\sin(0)}{x-0}=\sin'(0)=\cos(0)=1$