lim(h→0) [f(2 + h) - f(2)]/h
= lim(h→0) f'(2 + h)
= f'(2)
= (sinx)' |_(x = 2)
= cosx |_(x = 2)
= cos(2)
= - 0.42 ,in radians
If f(x) = sin x, evaluate lim f(2+h) - f(2) / h h->0 Evaluat
2个回答
相关问题
-
证明lim( h→0)[f(x0 h) f(x0-h)-2f(x0)]/h2=f''(x0)
-
f（x)=x^2 求lim{f(x+h)-f(x)}/h h趋向于0
-
若f'(0)=2,求lim(h→0)f(x0-h)-f(x0)/2h的值
-
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0)
-
f(x+y)=f(x)+f(y)+2xy lim f(h)/h =0 h 趋向0 问f'（x) 和f(x)
-
f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0
-
设函数f(x)在x=x0处可导,则lim(h>0)[f(x0)-f(x0-2h)]/h
-
设f(x)在x=x0的临近有连续的2阶导数,证明：lim(h趋近0）f(x0+h)-f(x0-h)-2f(x0)/h^2
-
f'(2)=1,则lim h→0 f(2-h)-f(2)/h=?
-
设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值