回復 1# thankyou 的帖子
第 1 題題目:設 \(f(a)=g(a)=1\) 且 \(f\,'(a)=g\,'(a)=2\),則 \(\displaystyle \lim_{h\to0}\frac{1}{h}\left[\frac{f(a+2h)}{g(a+3h)}-\frac{f(a)}{g(a)}\right]=\)?
解答:
\(\displaystyle \lim_{h\to0}\frac{1}{h}\left[\frac{f(a+2h)}{g(a+3h)}-\frac{f(a)}{g(a)}\right]=\lim_{h\to0}\frac{1}{h}\left[\frac{f(a+2h)}{g(a+3h)}-\frac{1}{1}\right]\)
\(\displaystyle =\lim_{h\to0}\frac{1}{h}\left[\frac{f(a+2h)-g(a+3h)}{g(a+3h)}\right]\)
\(\displaystyle =\lim_{h\to0}\frac{1}{h}\left[\frac{\left(f(a+2h)-1\right)-\left(g(a+3h)-1\right)}{g(a+3h)}\right]\)
\(\displaystyle =\lim_{h\to0}\left[\frac{\left(f(a+2h)-1\right)}{h}-\frac{g(a+3h)-1}{h}\right]\frac{1}{g(a+3h)}\)
\(\displaystyle =\lim_{h\to0}\left[2\cdot\frac{\left(f(a+2h)-f(a)\right)}{2h}-3\cdot\frac{g(a+3h)-g(a)}{3h}\right]\frac{1}{g(a+3h)}\)
\(\displaystyle =\left[2\cdot f\,'(a)-3\cdot g\,'(a)\right]\frac{1}{g(a)}\)
\(=-2\)