หาลิมิตโดยโลปิตาล

$$ \lim_{x \to \infty } (1+\frac{2}{x} + \frac{3}{x^2}) ^x $$
วิธีทำ
\begin{align} \cssId{Step1}{ ให้ y = (1+\frac{2}{x} + \frac{3}{x^2}) ^x } \\ \cssId{Step2}{ \ln y = x \ln (1+\frac{2}{x} + \frac{3}{x^2}) } \\ \cssId{Step3}{ \lim_{x \to \infty } \ln y } & \cssId{Step4}{= \lim_{x \to \infty } x \ln (1+\frac{2}{x} + \frac{3}{x^2}) } \\ & \cssId{Step5}{ =\lim_{x \to \infty } \frac{\ln (1+\frac{2}{x} + \frac{3}{x^2})}{\frac{1}{x}} \qquad อยู่ในรูป \qquad \frac{0}{0}} \\ & \cssId{Step6}{= \lim_{x \to \infty } ( \frac{1}{1+\frac{2}{x}+\frac{3}{x^2}} )( \frac{-\frac{2}{x^2}-\frac{6}{x^3}}{-\frac{1}{x^2}} ) } \\ &\cssId{Step7}{ = \lim_{x \to \infty } ( \frac{1}{1+\frac{2}{x}+\frac{3}{x^2}} ) (2+\frac{6}{x}) } \\ &\cssId{Step8}{= 2 } \\ \cssId{Step9} {ดังนั้น \lim_{x \to \infty} y \quad = e^2 } \\ \end{align}