Two important limits
The first one:
lim(sinx/x), when x tends to 0, its value is 1.
The second one:
lim(1+1/x)^x, when x tends to positive infinity, its value is e.
Pinch criterion
If the functions a(x), b(x), c(x)
satisfy:
a(x)<=b(x)<=c(x)
When
x tends to x0
or x tends to infinity
At this time, both a(x) and c(x) have limits, and the limits are both A. Then again because: a(x)<=b(x)<=c(x), it is
deduced that b(x) is in
The limit when x tends to x0
or x tends to infinity
is also A.
Let's prove the first important limit:
lim(sinx/x) When x tends to 0, the limit value is 1.
As shown
in the picture : the above picture shows a quarter circle, the radius of the circle is 1.
If the angle AOB is x, the unit is radians.
sinx=CB
x=arc AB, here we need to know the arc length calculation formula, which is equal to radius * radian of the sector,
tanx=AD/OA=AD, where OA=radius=1
can be obtained from the figure:
sinx<=x< =tanx
x<=tanx see the figure below, and the arc AB>AB>sinx is obvious from the figure.
Divide both sides of the inequality by sinx at the same time to get:
1<=x/sinx<=tanx/sinx
which is:
1<=x/sinx<=1/cosx
This is still a certain distance away from our sinx/x limit:
we will wait Count down all the formulas to get:
cosx<=sinx/x<=1
If we can deduce that the limit of cosx is 1 when x tends to 0, then according to the clamping criterion, it can be derived: sinx/x, the limit when x tends to 0 is also 1.
ok, the problem is transformed into, find the limit of cosx at x->0.
In the interval from 0 to pi/2:
0<=|cosx-1|=1-cosx=2[sin(x/2)]^2
According to the above graph, x is in the range of 0 to pi/2, sinx<=x<=tanx
and x/2, then the range is further reduced. At this time, in the range of 0 to pi/4, sin(x /2)<=x/2<=tan(x/2)
At this time:
0<=|cosx-1|=1-cosx=2[sin(x/2)]^2 <=2*(x/ 2) (x/2)=x x/2
When x tends to 0, the limit of x*x/2 is 0,
so we get: when x tends to 0, the limit is 1.
cosx<=sinx/x<=1
Therefore, when sinx/x tends to 0, the limit is 1.
Rule 2:
Monotone bounded sequence must have a limit.
The proof is omitted. In Advanced Mathematics, Tongji University Edition, the proof is omitted and only geometric explanations are given.
The second important limit
lim(1+1/x)^x, when x tends to positive infinity, its value is e.
The proof is divided into two steps.
If you consider the case of positive integers.
Xn = (1+1/n)^n
1) According to Newton's binomial expansion, we expand the cases of Xn and Xn+1 respectively, and we can get that Xn+1 is greater than Xn. The description is monotonically increasing.
2) Then prove that it is bounded. For
example, expand Xn=(1+1/n)^n:
Since 1-1/n is less than 1, and (1-1/n)(1-2/n) <1*1
and so on:
less than
At this time, it shows that the number Xn is bounded, ok, which means lim(1+1/x)^x, when x tends to positive infinity, there is a limit, usually represented by the letter e.