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the question I am attempting is:

Show $f '(0) = 0$ for:

$$f(x) = \left\{ \begin{array}{lr} \frac{\sin(x)}{x} & : x \neq 0\\ 1 & : x=0 \end{array} \right.$$

So I got stuck after the following working:

$$ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\ = \lim_{h \to 0} \frac{\frac{\sin(h)}{h}-\frac{\sin(0)}{0}}{h}\\ = \lim_{h \to 0} \frac{\sin(h)}{h^2} $$ and the above limit does not exist. Am I not applying the derivative definition right? or am I going wrong somewhere else?

EDIT: I now see that f(0) = 1 from the definition of the function, however I still have: $$\lim_{h \to 0}\frac{\sin(h)-h}{h^2}$$ which I am struggling to evaluate. Splitting into two fractions the first fraction is still the one from above which does not exist.

Also this must be from definition of derivative directly and not by L'Hopitals rule as it has not been taught yet.

marineabcd
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3 Answers3

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Note that in the usual proof of the limit $\lim_{h\to 0} \frac{\sin(h)}{h} = 1$, it is shown that the inequality $$\cos(h) \le \frac{\sin(h)}{h} \le 1$$ holds for all $h$ in an interval around $0$. Assuming this result, we have $$\cos(h) - 1 \le \frac{\sin{h}}{h} - 1 \le 0$$ in this interval. Taking absolute value and multiplying by $|1/h|$ then yields $$0 \le \left|\frac{\sin{h}-h}{h^2}\right| \le \left|\frac{\cos(h)-1}{h}\right|$$ for all $h$ in an interval around $0$. The desired limit is then $0$ by the squeeze theorem and the standard trig limit $\lim_{x\to 0} \frac{\cos(x)-1}{x} = 0$.

Rolf Hoyer
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Your issue is in evaluating $f(0)$ as $(\sin{0})/0$: you haven't used the definition you have given, which has $f(0)=1$.

Then you have $$ \lim_{h \to 0} \frac{\frac{\sin{h}}{h}-1}{h} = \lim_{h \to 0} \frac{\sin{h}-h}{h^2} = \lim_{h\to 0} \frac{\cos{h}-1}{2h} = \lim_{h \to 0} \frac{-\sin{h}}{2} = 0, $$ using L'Hôpital's rule twice.

To avoid using L'Hôpital, the easiest way is to use the squeeze theorem: for $0<x<\pi/2$, $$ \sin{x}<x<\tan{x}, \tag{1} $$ which can be shown by drawing triangles inside and tangent to the unit circle. Dividing by $\sin{x}$ gives $$ 1<\frac{x}{\sin{x}}<\frac{1}{\cos{x}}, $$ which shows, taking $x$ to $0$, that the limit of $x/\sin{x}$ must be $1$. The difficult bit is now to show that the limit of $\frac{\sin{h}-h}{h^2}$ is actually zero.

The inequality (1) also shows that $$ 0<1-\frac{\sin{h}}{h}<\frac{\tan{h}-\sin{h}}{h} = \frac{\sin{h}}{h}\frac{1-\cos{h}}{\cos{h}} $$ Using the identity $1-\cos{h}=2\sin^2{(h/2)}$, and dividing by $h$, we have $$ 0<\frac{h-\sin{h}}{h^2} < \frac{\sin{h}}{h} \frac{2\sin{(h/2)}}{h} \frac{1}{\cos{h}} (2\sin{(h/2)}) $$ We know enough limits to compute the right-hand side: the first 3 factors all tend to $1$, and the last one tends to zero, so another sandwich theorem application gives $$ \lim_{h \to 0} \frac{\sin{h}-h}{h^2} =0. $$

I'm quite surprised that this can actually be pushed this far.

Chappers
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  • ahh thank you I see why the sin(0)/0 is not 0 now, though at this point in the module we haven't learned L'Hopitals rule yet so cant use it. Is there another way to proceed? – marineabcd Apr 09 '15 at 17:17
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    (-1). This is pretty much circular reasoning. L'Hopital rule is based on the knowing the derivative of the sine function. The OP was almost essentially asking to compute the limit from first principles. – Nicolas Bourbaki Apr 09 '15 at 17:18
  • @NicolasBourbaki Well, you need the limit of the sine function to be able to check that this function is continuous, so that asking about the derivative even makes sense. – Chappers Apr 09 '15 at 17:37
  • @Chappers You do not need to know the function is continuous before you try to differentiate it at a point. The notion of differentiability does not assume continuity. It is true that continuity is a necessary condition, but look at the definition used, no continuity assumption necessary. – Nicolas Bourbaki Apr 09 '15 at 17:55
  • And I've never seen a textbook introduce trigonometrical functions before L'Hôpital's rule. – Chappers Apr 09 '15 at 18:25
  • @Chappers How about "A First Course in Calculus" by Serge Lang, in fact there is no mention of L'Hopital rule anywhere? Or how about "Calculus" by Spivak. Apostol actually does integration before differentiation. Even a baby-book like Stewart introduces trigonometric functions and their derivatives before L'Hopital rule. – Nicolas Bourbaki Apr 09 '15 at 19:29
  • Spivak, at least, does L'Hôpital first, and then does the trigonometrical functions in terms of an area integral for the circle. Indeed, this exercise is in the chapter, with the suggestion that L'Hôpital be used. You might perhaps be more careful in accusations of circular reasoning without knowledge of the premises used. – Chappers Apr 09 '15 at 20:14
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In this answer, it is shown that for $0\le x\le\frac\pi2$, $$ \cos(x)\le\frac{\sin(x)}x\le1\tag{1} $$ So, for $0\lt x\le\frac\pi2$, and symmetrically for $-\frac\pi2\le x\lt0$, subtract $1$ from $(1)$ and divide by $x$: $$ 0=\lim_{x\to0}\frac{\cos(x)-1}{x}\le\lim_{x\to0}\frac{\frac{\sin(x)}x-1}{x-0}\le0\tag{2} $$ We have the leftmost equality because $$ \begin{align} \lim_{x\to0}\frac{1-\cos(x)}x &=\lim_{x\to0}\frac1x\frac{\sin^2(x)}{1+\cos(x)}\\ &=\lim_{x\to0}\frac{\sin(x)}x\lim_{x\to0}\frac{\sin(x)}{1+\cos(x)}\\[6pt] &=1\cdot0\tag{3} \end{align} $$ Therefore, by the Squeeze Theorem, $(2)$ says $$ f'(0)=0\tag{4} $$

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