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Several limit problems lin (x^3-3x+2)/(x-1)^2= x->+∞ lin [1/x(sinx)+xsin(1/x)]= x->+∞ Please give the solution process, the more detailed the better, and teach how to find the limit such as 0 / 0 and ∞ / ∞
It's easy to use Robita's law
For 0 / 0, ∞ / ∞, first derive the numerator and denominator respectively until the limit can be found
lin (x^3-3x+2)/(x-1)^2= lin(3x^2-3)/2(x-1)=lin(6x)/2=+∞
x->+∞ x->+∞ x->+∞
lin 1/x(sinx)+xsin(1/x)]= lin 1/x(sinx)+lin xsin(1/x)
x->+∞ x->+∞ x->+∞
=0+lin((sin(1/x))/(1/x))=0+1=1
x->+∞
Because | 1 / X (SiNx) | = 0, the limit of the former is 0,
Several limit problems 1. When x approaches 0, LIM (x ^ 3 * sin1 / x) / 1-cos ^ 2 (x)= 2. When x approaches infinity, limcosx / e ^ x + e ^ (- x)=
1. It can be transformed into a bounded function multiplied by infinitesimal. The result is 0
2. Same as 1. The result is 0
1 / (SiNx) ^ 2 - 1 / x ^ 2 when x tends to the limit of 0
The answer on the first floor is based on the popular Infinitesimal Substitution in China, which is ridiculously wrong. The "equal order infinitesimal substitution" that desperately misleads students is the most absurd example in domestic college teaching. Where can students who kill chickens with ox knives and learn the basic limit on the second floor learn the Taylor formula? It's like asking students from grade one and grade two to solve equations
Why does Lim1 / (x-1) have a limit when x approaches 1 Why does Lim1 / (x + 1) limit not exist when x approaches - 1? Why?
The limit should not exist
X approaches 0 Lim 1 / X (1 / sinx-1 / TaNx)
Original formula = LIM (x → 0) (SiNx / cosx SiNx) / (xsinxtanx)
=lim(x→0)(1-cosx)/(xtanx)*1/cosx
=lim(x→0)2sin^2(x/2)/(xtanx)
=lim(x→0)2*(x/2)^2/x^2 (x~sinx~tanx)
=1/2
lim 2^n *sin(x/2^n) n→∞ Find the limit
Let t = 2 ^ n, then t →∞
lim sin(x/t)/(1/t)
Equivalent infinitesimal
=lim x/t/（1/t）
=x
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