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Find LIM (x → 0) (SiNx / x) ^ (cosx / 1-cosx)
y=(sinx/x)^(cosx/1-cosx)
lny=(cosx(lnsinx-lnx)/(1-cosx)
limlny=lim(cosx(lnsinx-lnx)/(1-cosx)=lim(lnsinx-lnx)/(1-cosx)=lim(cosx/sinx-1/x)/sinx=lim(xcosx-sinx)/x^3=lim(cosx-xsinx-cosx)/3x^2=-1/3
limy=e^(-1/3)
Limit LIM (x-sinx) / [x (1-cosx)] where x tends to 0
lim(x→0)(x-sinx)/[x(1-cosx)]
=LIM (x → 0) (1-cosx) / [(1-cosx) + xsinx] Robita's law
=lim(x→0)sinx/[sinx+sinx+xcosx]
=lim(x→0)sinx/[2sinx+xcosx]
=lim(x→0)1/[2+xcosx/sinx]
=lim(x→0)1/lim(x→0)[2+xcosx/sinx]
=1/[2+1]
=1/3
Additional notes:
lim(x→0)xcosx/sinx
=lim(x→0)[cosx-xsinx]/cosx
=[1-0]/1
=1
If the limit of a function is positive infinity, is there no limit? For example, if the limit of a function is positive infinity, is it that there is no limit?
Yes, if the limit is infinite, then it is defined as that the limit does not exist. However, if the limit is infinitesimal, then the limit exists
High number limit operation Limf (x) = + 00 limg (x) = + 00 limh (x) = a why Are LIM (f (x) + G (x)) = + 00 and lim (f (x) + H (x)) = + 00 LIM (f (x) g (x)) = + 00 correct? Not according to the limit algorithm, what can't you do when infinity or limit doesn't exist?
In the question, "according to the limit algorithm, you can't do this when infinity or limit doesn't exist"
It is also true that "the operation criterion of limits requires that limits exist, and they are equal to + ∞ no"
LIM (f (x) + G (x)) = + ∞ and lim (f (x) + H (x)) = + ∞ and lim (f (x) g (x)) = + ∞ are indeed correct,
Their correct basis can be proved not according to the limit algorithm, but according to the definition of limit
High number limit operation Lim {1 / (1-x) -- 1 / (1-x ^ 3)}, X tends to 1
General points
lim[1/(1-x)--1/(1-x^3)]
=lim[(1+x +x ²- 1）/(1-x ³)]
=lim[(x+x ²））/ (1-x ³)]
=∞
Let function y = f (x) be a function defined on [- 1,1], then the intersection of the definition field of function f (x + 1) and f (x) + 1 is
Obviously, the domain of F (x) + 1 is still [- 1,1], which has nothing to say
F (x + 1) can lead to Z = x + 1. Because f (z) is defined in [- 1,1], the definition field of X is [- 2,0]
【-1,1】∩【-2,0】=【-1,0】
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