lim(x→0)sinx/x= 1
lim(x→∞)( 1+ 1/x)^x=e
1.lim) function is still infinitesimal, and the limit is 0).
2. lim[ root number (x+ 1)- root number (x- 1)]cosx
x→+∞
=0
(same as 1)
3.lim (sin2x-tanx)/x
x→0
= lim(x→0)[2 sin2x/(2x)-sinx/x *( 1/cosx)]
=2- 1
= 1
Forest [1-(cos 2) x]/x 2
x→0
=lim(x→0)(sinx/x)^2
= 1
5.lim (x-sinx)/(x+sinx)
x→0
= lim(x→0)[( 1-sinx/x)/( 1+sinx/x)
=0
6.lim nsin(π/n)
n→∞
=lim(n→∞)π*sin(π/n)/(π/n)
=π
(n→∞, then π/n→0)
7.lim arcsinx/3x
x→0
=lim(x→0) 1/[√( 1-x^2)*3]
= 1/3
(L'H?pital's law can be used, namely lim(x→0)(arcsinx)'/(3x)').
8. forest (tanx-sinx)/x 3
x→0
=lim(x→0)(sinx/cosx-sinx)/x^3
=lim(x→0)(sinx-sinx*cosx)/(x^3*cosx)
=lim(x→0){sinx-sinx*{ 1-2[sin(x/2)]^2}}/(x^3*cosx)
=lim(x→0)2sinx*[sin(x/2)]^2/(x^3*cosx)
=lim(x→0)[sinx/x*[sinx(x/2)/(x/2)]^2* 1/(2cosx)]
= 1/2
(chemical type (1+ 1/n) n)
9.lim sinx/(x-π)
x→π
=lim(x→π)sin(π-x)/(x-π)
=lim(x→π)-sin(x-π)/(x-π)
=- 1
lim[(x^2+ 1)/(x^2- 1)]^x^2
x→∞
=lim(x→∞)[ 1+2/(x^2- 1)]^(x^2)
=lim(x→∞){{ 1+ 1/[(x^2- 1)/2]}^[(x^2- 1)/2]}^2*[ 1+2/(x^2- 1)]
=e^2* 1
=e^2
1 1 . lim[(x-3)/(x+2)]^(2x+ 1)
x→∞
=lim(x→∞)[ 1-5/(x+2)]^(2x+ 1)
=lim(x→∞){{ 1+ 1/[-(x+2)/5]}^[-(x+2)/5]}^(- 10)*{[ 1-5/(x+2)]^(-3)}
=e^(- 10)* 1^(-3)
=e^(- 10)
12.lim (1-1/x) root number x
x→+∞
=lim(x→+∞){[ 1+ 1/(-x)]^(-x)}^(- 1/√x)
=e^0
= 1
Questions 10, 1 1, 12 should all be changed to (1+1/n) type.
(The formula is so long ...)
13.lim ln( 1+x)/x
x→0
=lim[ln( 1+x)]'/x '
= lim(x→0)[ 1/( 1+x)]/ 1
= 1
14 . lim ln( 1+3x)/sin2x
x→0
= lim(x→0)[3/( 1+3x)]/(2 cos 2x)
=3/2
15.lim n[ln(n+2)-lnn]
n→∞
= lim(n→∞)ln[(n+2)/n]/( 1/n)
=lim(n→∞)[n/(n+2)*(-2/n^2)]/(- 1/n^2)
=2
13, 14, 15 should use L'H?pital's law.
16.lim radical number (1-x)-3/x+8
x→-∞
=lim(x→-∞)( 1-x-3^2)/[(x+8)(√( 1-x)+3)]
= lim(x→-∞)[- 1/(√( 1-x)+3)]
=0
I don't know if I understand this formula correctly, but the numerator should be a rational number (this problem is multiplied by [√( 1-x)+3)].
17 . lim【x/( 1+x)]^(-2x+ 1
x→∞
=lim(x→∞){[ 1- 1/( 1+x)]^(- 1-x)}^2*[x/( 1+x)]^3
=e^2
18 . lim【sin(9x^2- 16)]/(3x-4
x→4/3
= forest (x → 4/3) [sin (9x 2-16)]/(9x 2-16) * (3x+4)
= 1*(3*4/3+4)
=8
lim [(x+h)^3-x^3]/h
h→0
=lim(h→0)(x^3+3x^2h+3xh^2+h^3-x^3)/h
=lim(h→0)(3x^2+3xh+h^2)
=3x^2
The Deduction Process of (x 3)' = 3x 2
20. lim[ radical number (1-x)-3]/[2+3 radical number x]
x→-∞
There is something wrong with this question. I calculated with software and got:1/2-√ 3/2 * i.
(People with imaginary numbers all say ...)
But the basic idea is √( 1+x)+3 at the same time twice, then the molecule is -x-8, which can be written as-[x (1/3)+2] [x (2/3)-2 * x (1/0).
2 1.lim (x+ 1)/[ root number (x+5)-2]
x→- 1
= lim(x→- 1)(x+ 1)[√( x+5)+2]/(x+5-4)
=lim(x→- 1)√(x+5)+2
=4
Molecular physics and chemistry
Some questions are not clearly written, such as16,20 questions.
Because it's too long, it's not convenient to see. If you have any questions you don't understand, please add them. ...
Are you sure about x→-∞? In that case ...
16、
The highest power term of x in the molecule is (1-X) 1/2, which means that the highest power of x is 1/2, and the highest power of x in the denominator is 1.
So lim (x → -∞) [√ (1-x)-3]/(x+8) = 0.
20、
The highest power of X in the numerator is 1/2, and the highest power of X in the denominator is 1/3.
So lim (x →-∞) [√ (1-x)-3]/[2+x (1/3)] = +∞
So there is no limit.