[Math] L’Hôpital’s rule does not apply?!

calculuslimits

Apparently, Rogawski's Calculus for AP contains the following problem:

108. Explain why L'Hôpital's rule does not apply to

$$ \lim_{x\rightarrow 0}\frac{x^2\sin\frac{1}{x}}{\sin x} $$

It seems to me that it does apply:

The L'Hôpital's rule says: if $\lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}g(x)=0$ and both $f$ and $g$ are differentiable at $x=c$ and $g'(c)\ne 0$, then $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}$ exists and is equal to $\frac{f'(c)}{g'(c)}$.
(Note that nothing is assumed about differentiability of $f$ and $g$ other than at $x=c$).

  1. Define the numerator $f(x)=x^2\sin\frac{1}{x}$ to be $f(0)=0$ at $x=0$. Now, both numerator $f$ and denominator $g(x)=\sin(x)$ are continuous at $x=0$ and their values are $f(0)=g(0)=0$.

  2. The numerator $f$ is differentiable at $x=0$ and the derivative is $f'(0)=0$ (the derivative itself is discontinuous at 0, but that is irrelevant – even the existence of the derivative at any point other than 0 does NOT matter). One can see that from the definition of the derivative:
    $f'(0)=\lim_{h\rightarrow 0} \frac{h^2\sin\frac{1}{h}}{h} =
    \lim_{x\rightarrow 0} h\sin\frac{1}{h} = 0$ (see PS step 2 below).

  3. The denominator $g$ is differentiable at $x=0$ and the derivative is $g'(0)=\cos 0=1$.

  4. Thus the limit is $\frac{0}{1} = 0$.

What am I missing?

PS. Note that I am not asking why the limit is 0. That can be easily seen without L'Hôpital:

  1. $\lim_{x\rightarrow 0}\frac{x}{\sin x} = 1$:
    this is the inverse of the standard limit $\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$.

  2. $\lim_{x\rightarrow 0} x \sin\frac{1}{x} = 0$
    because $\sin\frac{1}{x}$ is bounded and $\lim_{x\rightarrow 0} x = 0$,
    this follows from Squeeze theorem.

  3. the Product Rule for Limits
    implies that
    $$\lim_{x\rightarrow 0}\frac{x^2\sin\frac{1}{x}}{\sin x} =
    \lim_{x\rightarrow 0}x\sin\frac{1}{x} \times
    \lim_{x\rightarrow 0}\frac{x}{\sin x} =
    0 \times 1 = 0$$

PPS Here is the scan from the textbook:

enter image description here

Best Answer

You have misstated L'Hopital's Rule. It does not say $\lim_{x\to c}{f(x)\over g(x)}={f'(c)\over g'(c)}$ (with the usual assumptions on $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$). It says

$$lim_{x\to c}{f(x)\over g(x)}=\lim_{x\to c}{f'(x)\over g'(x)}$$

provided the latter limit exists. In this case

$${f'(x)\over g'(x)}={2x\sin(1/x)-\cos(1/x)\over\cos x}$$

for $x\not=0$. So even though $f'(0)=\lim_{x\to0}{f(x)-f(0)\over x}=\lim_{x\to0}x\sin(1/x)=0$ (assuming we let $f(0)=\lim_{x\to0}f(x)=0$), the hypotheses of L'Hopital's Rule are not fulfilled because $\lim_{x\to0}(f'(x)/g'(x))$ does not exist. In particular $\cos(1/x)$ has no limit as $x\to0$.

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