limx-1/x ,求x趋近于1的极限?limπ(x-1)/sin（πx） ,求x趋近于1的极限?π是派~

lim_(x->1) (x-1/x)
The limit of a difference is the difference of the limits:
= lim_(x->1) x-lim_(x->1) 1/x
The limit of a quotient is the quotient of the limits:
= lim_(x->1) x-1/(lim_(x->1) x)
The limit of x as x approaches 1 is 1:
= 0
lim_(x->1) (pi (x-1))/(sin(pi x))
Factor out constants:
= pi (lim_(x->1) (x-1)/(sin(pi x)))
Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->1) (-1+x)/(sin(pi x)) = lim_(x->1) (( d(-1+x))/( dx))/(( dsin(pi x))/( dx)):
= pi (lim_(x->1) 1/(pi cos(pi x)))
Factor out constants:
= lim_(x->1) 1/(cos(pi x))
The limit of a quotient is the quotient of the limits:
= 1/(lim_(x->1) cos(pi x))
Using the continuity of cos(x) at x = pi write lim_(x->1) cos(pi x) as cos(lim_(x->1) pi x):
= 1/(cos(lim_(x->1) pi x))
Factor out constants:
= 1/(cos(pi (lim_(x->1) x)))
The limit of x as x approaches 1 is 1:
= -1

你好！
前面的分子直接等于0，而分母是1，所以极限是0
后面的题目用罗比达法则求导做，结果是-1

limx-1/x=0
lim(x→1)π(x-1)/sin（πx）(0/0型,运用洛必达法则上下求导得）
=lim(x→1)π/[πcos（πx）]
=-1