Exercise 5.3.10

Question

Let $f(x)=x \sin \left(1 / x^{4}\right) e^{-1 / x^{2}}$ and $g(x)=e^{-1 / x^{2}}$. Using the familiar properties of these functions, compute the limit as $x$ approaches zero of $f(x), g(x), f(x) / g(x)$, and $f^{\prime}(x) / g^{\prime}(x)$. Explain why the results are surprising but not in conflict with the content of Theorem 5.3.6. ${ }^{1}$

Ulisse Mini · Accepted Answer

$$
  \lim_{x 	o 0} f(x) = 0,
  \quad \lim_{x 	o 0} g(x) = 0,
  \quad \lim_{x 	o 0} \frac{f(x)}{g(x)} = 0,
  $$
  To compute the last limit we need to find $f'$ and $g'$ using derivative rules. Let $h(x) = x\sin(1/x^4)$ for bookkeeping.
  $$
  \begin{aligned}
    g'(x) &= \frac{d}{dx} e^{-x^{-2}} = e^{-x^{-2}}(2x^{-3}) = g(x)(2x^{-3}) \
    h'(x) &= \sin(x^{-4}) + x\cos(x^{-4})(-4x^{-5}) = \sin(x^{-4}) - 4x^{-4}\cos(x^{-4})\
    f'(x) &= \frac{d}{dx} h(x)g(x) \
          &= h'(x)g(x) + g(x)(2x^{-3})h(x) \
          &= g(x)(h'(x) + 2x^{-3}h(x))
  \end{aligned}
  $$
  We can simplify by dividing out $g(x)$
  $$
  \frac{f'(x)}{g'(x)}
  = \frac{g(x)(h'(x) + 2x^{-3}h(x))}{g(x)(2x^{-3})}
  = \frac{h'(x) + 2x^{-3}h(x)}{2x^{-3}}
  = \frac 12 x^3h'(x) + h(x)
  $$
  Now we can compute the limit
  $$
  \begin{aligned}
  \lim_{x 	o 0}\left[\frac 12 x^3 h'(x) + h(x)ight]
  &= \lim_{x 	o 0} \frac 12 x^3 h'(x) &&{	ext{since } \lim_{x	o 0} h(x) = 0} \
  &= \frac 12 \lim_{x 	o 0}\left[x^3\sin(x^{-4}) - 4x^{-1}\cos(x^{-4})ight] &&{	ext{substitute } h'} \
  &= -2 \lim_{x 	o 0} x^{-1}\cos(x^{-4}) &&{	ext{since } \lim_{x 	o 0} x^3\sin(x^{-4}) = 0} \
  &	o 	ext{does not exist, just like } \lim_{x 	o 0} 1/x
  \end{aligned}
  $$
  L'Hopital's rule for $0/0$ would apply if $f'/g'$ existed. But when it doesn't exist $f/g$ may still exist. Put another way, the converse of L'Hopital's rule does not hold.

Exercise 5.3.10

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Exercise 5.3.10

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