陶哲轩实分析（上）10.5及习题-Analysis I 10.5

著名的洛必达法则。

Exercise 10.5.1

By Proposition 10.1.7, since $g' (x_0 )≠0$, we can find a $δ>0$ such that for all $x∈X$ and $|x-x_0 |≤δ$, we have
$|g(x)-g(x_0 )-g' (x_0 )(x-x_0 )|≤\frac{|g' (x_0 )|}{2} |x-x_0 |$
Using triangle inequality and $g(x_0 )=0$, we further get
$\frac{|g' (x_0 )|}{2} |x-x_0 |≤|g(x)|$
Thus if $x∈\left(X∩(x_0-δ,x_0+δ)\right)-\{x_0 \}$, we shall have $|g(x)|>0$, or $g(x)≠0$. In this case we can safely use Proposition 9.3.14 to have
$\lim_{x→x_0;x∈(X∩(x_0-δ,x_0+δ))-\{x_0 \} } ⁡\frac{f(x)}{g(x)}=\lim_{x→x_0;x∈(X∩(x_0-δ,x_0+δ))-\{x_0 \} } \frac{f(x)-f(x_0)}{g(x)-g(x_0 )}\\=\frac{\lim_{x→x_0;x∈(X∩(x_0-δ,x_0+δ))-\{x_0 \} }⁡\frac{f(x)-f(x_0 )}{x-x_0 }}{\lim_{x→x_0;x∈(X∩(x_0-δ,x_0+δ))-\{x_0 \} }⁡\frac{g(x)-g(x_0)}{x-x_0 }} =\frac{(f' (x_0 )}{g' (x_0 ) }$

Exercise 10.5.2

In the first example, we have $g(x)=1+x↛0$ if $x→0$.
In the second example, notice that to use the L’Hôpital’s rule, we need first the limit
$\lim_{x→a;x∈(a,b]}⁡\frac{f' (x)}{g' (x) }$
exist, then we can safely use the rule, but in this example, the limit
$\lim_{x→0+;x∈(0,∞) }⁡\frac{(x^2 \sin⁡(x^{-4} )) '}{x'}$
doesn’t exist, due to the calculations in the textbook. So it’s not safe to use the L’Hôpital’s rule.