题目
已知函数f(x)在点x0处可导,则下列极 限 中 () 等于导数值f`(x0).-|||-(A) lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0))}(h) (B) lim _(harrow 0)dfrac (f({x)_(0)-3h)-f((x)_(0))}(h)-|||-(C) lim _(harrow 0)dfrac (f({x)_(0))-f((x)_(0)-h)}(h) (D) lim _(harrow 0)dfrac (f({x)_(0))-f((x)_(0)+h)}(h)
题目解答
答案
解析
步骤 1：分析选项 (A)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0}+2h)-f({x}_{0})}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0}+2h)-f({x}_{0})}{2h} \times 2$
= $2f'({x}_{0})$
步骤 2：分析选项 (B)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0}-3h)-f({x}_{0})}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0}-3h)-f({x}_{0})}{-3h} \times (-3)$
= $-3f'({x}_{0})$
步骤 3：分析选项 (C)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}-h)}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}-h)}{-h} \times (-1)$
= $f'({x}_{0})$
步骤 4：分析选项 (D)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}+h)}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}+h)}{h}$
= $-f'({x}_{0})$
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0}+2h)-f({x}_{0})}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0}+2h)-f({x}_{0})}{2h} \times 2$
= $2f'({x}_{0})$
步骤 2：分析选项 (B)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0}-3h)-f({x}_{0})}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0}-3h)-f({x}_{0})}{-3h} \times (-3)$
= $-3f'({x}_{0})$
步骤 3：分析选项 (C)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}-h)}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}-h)}{-h} \times (-1)$
= $f'({x}_{0})$
步骤 4：分析选项 (D)
$\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}+h)}{h}$
= $\lim _{h\rightarrow 0}\dfrac {f({x}_{0})-f({x}_{0}+h)}{h}$
= $-f'({x}_{0})$