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)