Prove this function is differentiable and find it's derivative. $$ S(x)= \begin{cases} \frac{\sin x}{x} & x\neq0\\ 1 & x=0 \end{cases} $$ The derivative in $\mathbb{R}-\{0\}$ is : $\frac{x\cos x-\sin x}{x^2}$, but I don't know how to find the derivative at $x=0$
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Just use the definition of the derivative in one point… what's the definition for a function to be differentiable in an arbitrary $x_0$? Then set $x_0 = 0$. Where are you stuck? – Gono Aug 08 '20 at 18:12
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$\lim _{h\to 0} \frac{S(h)-S(0)}{h}$ – user-492177 Aug 08 '20 at 18:14
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Yes, what is is $S(h)$, what is $S(0)$? Where are you stuck in calculating this limit? – Gono Aug 08 '20 at 18:17
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@Gono I get to $\lim_{h\to 0}(\sin h - h)/h^2$ and I don't know where to go next. – Aug 08 '20 at 18:23
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1Now try using L Hopital to evaluate the limit... – sai-kartik Aug 08 '20 at 18:23
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@Wrench Yes. Thanks. – Aug 08 '20 at 18:42
3 Answers
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$$\frac{\frac{\sin x}{x}-1}{x} = \frac{\sin x -x}{x^2}= \frac{3\sin \frac{x}{3}-4 \sin^3 \frac{x}{3}-x}{x^2} \sim \frac{3\sin \frac{x}{3}-x}{x^2}=3\frac{\sin \frac{x}{3}-\frac{x}{3}}{x^2}$$ Second and last member without $3$ have same limits. Can you continue?
zkutch
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Okay, you might do something like this
$$ f’(0) = \lim_{h\to 0} \frac{f(h) - f(0)}{h}\\ f’(0)= \lim_{h\to 0} \frac{\frac{\sin h }{h} -1}{h} $$
and then use L’Hospitals rule to evaluate the limit. But what you forget is that the given function is discontinuous at zero and hence derivative will not exist at $x=0$.
Knight wants Loong back
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Since $S$ is even, its derivative at $x=0$ is either $0$ or undefined. There are many ways to show the former holds here, e.g.$$\lim_{x\to0}\frac{x\cos x-\sin x}{x^2}=\lim_{x\to0}\frac{\cos x-1}{x}+\lim_{x\to0}\frac{x-\sin x}{x^2}=0+0=0.$$Indeed$$\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12,\,\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$$are famous results which, multiplied by $\lim_{x\to0}x=0$, gives what we need.
J.G.
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