洛必达法则

洛必達法則可以求出特定函數趨近於某數的極限值。令${\displaystyle c\in {\bar {\mathbb {R} }}}$（擴展實數），兩函數${\displaystyle f(x),g(x)}$在以${\displaystyle x=c}$為端點的開區間可微，${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}\in {\bar {\mathbb {R} }}}$，並且${\displaystyle g'(x)\neq 0}$。

如果 ${\displaystyle \lim _{x\to c}{f(x)}=\lim _{x\to c}{g(x)}=0}$ 或 ${\displaystyle \lim _{x\to c}{|f(x)|}=\lim _{x\to c}{|g(x)|}=\infty }$ 其中一者成立，則稱欲求的極限${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}$為未定式。

此時洛必达法则表明：

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$。

對於不符合上述分數形式的未定式，可以通過運算轉為分數形式，再以本法則求其值。以下列出數例：

欲求的極限	條件	轉換為分數形式的方法
（1）${\displaystyle \lim _{x\to c}f(x)g(x)\!}$	${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{1/g(x)}}\!}$ 或 ${\displaystyle \lim _{x\to c}{\frac {g(x)}{1/f(x)}}\!}$
（2）${\displaystyle \lim _{x\to c}(f(x)-g(x))\!}$	${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}(f(x)-g(x))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)g(x))}}\!}$
（3）${\displaystyle \lim _{x\to c}{f(x)}^{g(x)}\!}$	${\displaystyle \lim _{x\to c}f(x)=0^{+},\lim _{x\to c}g(x)=0\!}$ 或 ${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=0\!}$	${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!}$
（4）${\displaystyle \lim _{x\to c}{f(x)}^{g(x)}\!}$	${\displaystyle \lim _{x\to c}f(x)=1,\ \lim _{x\to c}g(x)=\infty \!}$	${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!}$

注意：不能在数列形式下直接用洛必達法則，因為對於離散變量是无法求导数的。但此时有形式类近的斯托尔兹－切萨罗定理（Stolz－Cesàro theorem）作为替代。

下面仅给出 ${\displaystyle \lim _{x\to a}{f(x)}=\lim _{x\to a}{g(x)}=0,\,g'(a)\neq 0}$ 的证明。

设两函數${\displaystyle f(x)}$及${\displaystyle g(x)}$在a 點附近连续可导，${\displaystyle \ f(x)}$及${\displaystyle \ g(x)}$都在 a 點連續，且其值皆為 0 ，

${\displaystyle f(a)=0;\;g(a)=0,\qquad \lim _{x\to a}f(x)=0;\;\lim _{x\to a}g(x)=0}$

为了叙述方便，假设两函数在 a 点附近都不为0。另一方面，两函数的导数比值在 a 点存在，记为

${\displaystyle \lim _{x\to a}{\frac {f'(x)}{g'(x)}}=L.}$

由極限的定义，对任何一个${\displaystyle \epsilon >0}$（試想像y軸），都存在${\displaystyle \eta >0}$（試想像x軸），使得对任意的${\displaystyle a-\eta \leqslant x\leqslant a+\eta ,\,\,x\neq a}$，都有：

${\displaystyle L-\epsilon \leqslant {\frac {f'(x)}{g'(x)}}\leqslant L+\epsilon }$

而根据柯西中值定理（逆定理），对任意的${\displaystyle a-\eta \leqslant x\leqslant a+\eta ,\,\,x\neq a}$，都存在一个介于${\displaystyle a}$和${\displaystyle x}$之间的数${\displaystyle \xi }$，使得：

${\displaystyle {\frac {f(x)}{g(x)}}}$	${\displaystyle ={\frac {f(x)-f(a)}{g(x)-g(a)}}={\frac {f'(\xi )}{g'(\xi )}}}$
于是，	${\displaystyle L-\epsilon \leqslant {\frac {f(x)}{g(x)}}\leqslant L+\epsilon }$

因此，

极限${\displaystyle \lim _{x\to a}{\frac {f(x)}{g(x)}}=L=\lim _{x\to a}{\frac {f'(x)}{g'(x)}}.}$

${\displaystyle \lim _{x\to 0}{\frac {\sin \pi x}{\pi x}}\,}$	${\displaystyle =\lim _{x\to 0}{\frac {\sin x}{x}}\,}$
	${\displaystyle =\lim _{x\to 0}{\frac {\cos x}{1}}\,}$	${\displaystyle ={\frac {1}{1}}=1\,}$

${\displaystyle \lim _{x\to 0}{2\sin x-\sin 2x \over x-\sin x}}$	${\displaystyle =\lim _{x\to 0}{2\cos x-2\cos 2x \over 1-\cos x}}$
	${\displaystyle =\lim _{x\to 0}{-2\sin x+4\sin 2x \over \sin x}}$
	${\displaystyle =\lim _{x\to 0}{-2\cos x+8\cos 2x \over \cos x}}$
	${\displaystyle ={-2\cos 0+8\cos 0 \over \cos 0}}$
	${\displaystyle =6\,}$

${\displaystyle \lim _{x\to 0}{r^{x}-1 \over x}}$	${\displaystyle =\lim _{x\to 0}{{\frac {d}{dx}}r^{x} \over {\frac {d}{dx}}x}}$
	${\displaystyle =\lim _{x\to 0}{r^{x}\ln r \over 1}}$
	${\displaystyle =\ln r\lim _{x\to 0}{r^{x}}}$
	${\displaystyle =\ln r\!}$

${\displaystyle \lim _{x\to 0}{e^{x}-1-x \over x^{2}}=\lim _{x\to 0}{e^{x}-1 \over 2x}=\lim _{x\to 0}{e^{x} \over 2}={1 \over 2}}$

${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {x}}{\ln(x)}}=\lim _{x\to \infty }{\frac {\ {\frac {1}{2{\sqrt {x}}}}\ }{\frac {1}{x}}}=\lim _{x\to \infty }{\frac {\sqrt {x}}{2}}=\infty }$

${\displaystyle \lim _{x\to \infty }x^{n}e^{-x}=\lim _{x\to \infty }{x^{n} \over e^{x}}=\lim _{x\to \infty }{nx^{n-1} \over e^{x}}=n\lim _{x\to \infty }{x^{n-1} \over e^{x}}=0}$

${\displaystyle \lim _{x\to 0+}(x\ln x)=\lim _{x\to 0+}{\ln x \over {\frac {1}{x}}}=\lim _{x\to 0+}{{\frac {1}{x}} \over -{\frac {1}{x^{2}}}}=\lim _{x\to 0+}-x=0}$

${\displaystyle \lim _{t\to 0}\,\mathrm {sinc} (f_{0}t)\cdot {\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$	${\displaystyle =\left\{\lim _{t\to 0}\,\mathrm {sinc} (f_{0}t)\right\}\cdot \left.{\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}\,\right|_{t=0}}$
	${\displaystyle =1\cdot 1=1}$

${\displaystyle \lim _{t\to {\frac {1}{2\alpha f_{0}}}}\mathrm {sinc} (f_{0}t)\cdot {\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$	${\displaystyle =\mathrm {sinc} \left({\frac {1}{2\alpha }}\right)\cdot \lim _{t\to {\frac {1}{2\alpha f_{0}}}}{\frac {\cos \left(\pi \alpha f_{0}t\right)}{\left[1-\left(2\alpha f_{0}t\right)^{2}\right]}}}$
	${\displaystyle =\mathrm {sinc} \left({\frac {1}{2\alpha }}\right)\cdot \left({\frac {\frac {-\pi }{2}}{-2}}\right)}$
	${\displaystyle =\sin \left({\frac {\pi }{2\alpha }}\right)\cdot {\frac {\alpha }{2}}}$

• 极限