The second is equivalent infinitesimal.
tanx~x
That's tanx~3x.
The result is.
The second diagram uses two important limits.
x-& gt; Infinity (1+1/x) x->; e
Original title: X-> 0 is1/x->; Infinite; unbordered
( 1+2x)^( 1/x)
=[( 1+ 1/( 1/2x))^ 1/2x]^2
The front just fits the formula.
The result is e 2.
The third picture
( 1- 1/x)^kx
=( 1+ 1/(-x))^[(-x)*(-k)]
=e^(-k)
The fourth picture is not clear.
The fifth picture
Using equivalent infinitesimal transformation
E x-1~x Sinks ~ x
So molecular change is
(x+ 1)-(x+ 1)
Denominator becomes x
It's two o'clock.
You can also use L'H?pital's law to derive the upper and lower derivatives respectively, as shown below.
(e^x+e^(-x))/cosx
In this era, if you enter 0, you can calculate 2.
The sixth picture
Find the integral,
Use component integration first.
(xe^-x)dx
=-xe^(-x)-[-e^(-x)dx]
=-xe^(-x)-e^(-x)
Then substitute 1 0.
The result is [-e (-1)-e (-1)]-[0-1] =1.
The Last Photograph
You can get the indefinite integral first.
x/(sinx)^2dx
=x*(cscx)^2dx
=-xdcotx
=-xcotx-(-cotxdx)
=-xcotx+cosx/sinxdx
=-xcotx- 1/sinxdsinx
=-xcotx-ln[sinx]
The replacement result is:
[-pi/3 * cot(pi/3)-ln[sinpi/3]]-[-pi/4 cot pi/4-ln[sinpi/4]]
=[-pi* radical 3/3-ln (radical 3/2)]-[-pi/4* 1-ln (radical 2/2)]
=( 1- root number 3/3) pi-(1/2ln3-ln2)+(1/2ln2-ln2)
=( 1- root number 3/3)pi- 1/2(ln3-ln2)