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!
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3 Answers
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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 $\displaystyle 1=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\sin(x)}{x}$.
answered Dec 9, 2012 at 18:34
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$\begingroup$ Don't I have to show the continuity of sin(x) and f(x)=1/x then? $\endgroup$– JoshDec 9, 2012 at 19:03
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1$\begingroup$ Yes, technically you do, but I am sure those are facts you can assume. It should have been part of your course that every polynomial is continuous, and that division of functions are continuous where the denominator is nonzero--this gives $\frac{1}{x}$. The fact that $\sin(x)$ is continuous is more delicate, and depends on how you define $\sin$. $\endgroup$ Dec 9, 2012 at 19:06
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$\begingroup$ Okay, got your point. But now, how to verify that $\displaystyle 1=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\sin(x)}{x}$? $\endgroup$– JoshDec 9, 2012 at 19:20
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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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1$\begingroup$ But don't you need to know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to prove that $\sin x$ is differentiable? $\endgroup$– JavierDec 9, 2012 at 18:42
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2$\begingroup$ Not if you define it in terms of power series, or $e^x$. How else would you define $\sin(x)$? $\endgroup$ Dec 9, 2012 at 18:48
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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:
$$\lim_{x → 0} \frac{\sin(x)}{x}=\lim_{x → 0} \frac{\sin(x)-\sin(0)}{x-0}=\sin'(0)=\cos(0)=1$$
