Prove or give a counterexample to the following assertion. Let \(f,g\) be real-valued \(C^1\) functions on \((0,1)\) such that \[ \lim_{x\to0}f(x)=\lim_{x\to0}g(x)=0, \] and suppose \(g\) and \(g'\) never vanish. If \[ \lim_{x\to0}\frac{f(x)}{g(x)}=c, \] must one have \[ \lim_{x\to0}\frac{f'(x)}{g'(x)}=c? \]
Solution
The assertion is false.
1. Define on \((0,1)\)\[ g(x)\coloneqq x, \qquad f(x)\coloneqq x^2\sin\!\left(\frac1{x^2}\right). \]Then \(f\) and \(g\) satisfy all hypotheses with\[ c=0. \]
Both functions are \(C^1\) on \((0,1)\). Moreover, \[ g(x)=x\ne0, \qquad g'(x)=1\ne0 \] throughout that interval. As \(x\to0^+\), \[ \abs{f(x)} \le x^2 \longrightarrow0, \] and \(g(x)\to0\). Finally, \[ \frac{f(x)}{g(x)} = x\sin\!\left(\frac1{x^2}\right), \] whose absolute value is at most \(x\). Hence \[ \lim_{x\to0^+}\frac{f(x)}{g(x)}=0. \]
2. For these functions,\[ \frac{f'(x)}{g'(x)} = 2x\sin\!\left(\frac1{x^2}\right) - \frac2x\cos\!\left(\frac1{x^2}\right), \]and this expression has no finite limit as \(x\to0^+\).
Differentiating gives the displayed formula because \(g'(x)=1\).
For \[ x_n\coloneqq\frac1{\sqrt{2\pi n}}, \] one has \[ \sin(1/x_n^2)=0, \qquad \cos(1/x_n^2)=1, \] so \[ \frac{f'(x_n)}{g'(x_n)} = -\frac2{x_n} \longrightarrow-\infty. \] For \[ y_n\coloneqq\frac1{\sqrt{(2n+1)\pi}}, \] one has \[ \sin(1/y_n^2)=0, \qquad \cos(1/y_n^2)=-1, \] so \[ \frac{f'(y_n)}{g'(y_n)} = \frac2{y_n} \longrightarrow+\infty. \] Thus the derivative quotient cannot converge.
3. Therefore the proposed converse to l’Hôpital’s rule fails.
Step 1 gives functions satisfying the assumptions with \(c=0\), while step 2 shows that \[ \frac{f'(x)}{g'(x)} \] does not tend to \(0\) or to any finite value.
4. Q.E.D.
Step 3 supplies the required counterexample.