1st lecture: keep practising
Then we turn to the big topic: differentiation!
(1) Using the squeezing theorem to explain/show that 
%20%3D%200#card=math&code=%5Clim_%7Bx%20%5Cto%200%7D%20%28x%20%5Csin%20%5Cfrac%7B1%7D%7Bx%7D%29%20%3D%200).
This is because of 
, since 
.
Compare with 
. (Do not mix these two!)
(2) Examples 2.18 to 2.20 (The so-called 
indefinite limits.)
Involving with rational functions (sometimes even with radical)
Master the trick of simplifying complex fractions, such as reduction to a common denominator.
(3) Example 2.21, evaluating one-sided limits by definition
(4) Infinite limits (for comprehension) P. 146
Functions with infinite limits at some point is somehow managable, not so bad as those functions without limits, for instance 
%20%3D%201#card=math&code=D%28x%29%20%3D%201) if 
is rational, %3D0#card=math&code=D%28x%29%3D0) if is irrational. (Dirichlet function)
Classical examples on infinite limits:
Convention: (
is finite)
, 
%20%3D%20%5Cpm%20%5Cinfty#card=math&code=a%20%5Ccdot%20%28%5Cpm%20%5Cinfty%29%20%3D%20%5Cpm%20%5Cinfty) (assume 
)
See also Theorem 2.3 in the OpenStax textbook.
Now we turn to differentiation.
Key Idea: approximation!
Example: Our earth has radius 6367 km (approximately). Suppose someday in the future the radius of the earth is increased by 50 cm. How much longer is equator of the earth in the future, compare to the equator of the earth now? How much larger is the volume of the earth in the future, compare to the volume of the earth now. (First image, then calculate your answer.)
Answer: The length of the equator = 
the radius of the earth.
So the length of the new equator is
#card=math&code=2%5Cpi%5Ctimes%20%286367%20km%20%2B%2050%20cm%29) . While the length of the current equator is 
#card=math&code=2%5Cpi%20%5Ctimes%20%286367%20km%29) .
So the increment of the equator is just 
.
And the volume of the earth: 
%5Cpi%20r%5E3#card=math&code=V%20%3D%20%284%2F3%29%5Cpi%20r%5E3), where 
is the radius of the earth.
The increment of the volume of the earth is:
= 
%5Cpi%20%5B(6367%20km%20%2B%2050%20cm)%5E3%20-%20(6367%20km)%5E3%5D#card=math&code=%284%2F3%29%5Cpi%20%5B%286367%20km%20%2B%2050%20cm%29%5E3%20-%20%286367%20km%29%5E3%5D).
Let 
. In unit: meter:
%5E3%3D(6%2C367%2C000%20%2B%200.5%20)%5E3%20%3D%20y%2Bdy#card=math&code=%28r%2Bdr%29%5E3%3D%286%2C367%2C000%20%2B%200.5%20%29%5E3%20%3D%20y%2Bdy) ,
in unit: m
So
%5Cpi%20dy%20%3D%20(4%2F3)%5Cpi%20%5Cfrac%7Bdy%7D%7Bdr%7D%20%5Ctimes%20dr%5C%5C%20%0A%0A%26%5Capprox%20%20(4%2F3)%5Cpi%20%5Ctimes%203%20%5Ctimes%20r%5E2%20%5Ctimes%20dr%5C%5C%0A%26%3D%204%5Cpi%20r%5E2%20%5Ccdot%20dr%20%5C%5C%0A%26%3D%202%5Cpi%20%5Ctimes%206367000%5E2%20(%5Cmathrm%7Bm%7D%5E3)%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0AV%7Bnew%7D-%20V%7Bcurrent%7D%20%26%3D%20%284%2F3%29%5Cpi%20dy%20%3D%20%284%2F3%29%5Cpi%20%5Cfrac%7Bdy%7D%7Bdr%7D%20%5Ctimes%20dr%5C%5C%20%0A%0A%26%5Capprox%20%20%284%2F3%29%5Cpi%20%5Ctimes%203%20%5Ctimes%20r%5E2%20%5Ctimes%20dr%5C%5C%0A%26%3D%204%5Cpi%20r%5E2%20%5Ccdot%20dr%20%5C%5C%0A%26%3D%202%5Cpi%20%5Ctimes%206367000%5E2%20%28%5Cmathrm%7Bm%7D%5E3%29%0A%5Cend%7Baligned%7D%0A)
That is, 
#card=math&code=V%7Bnew%7D-%20V%7Bcurrent%7D%20%5Capprox%202%5Cpi%20%5Ctimes%206367%20%28%5Cmathrm%7Bkm%7D%5E3%29).
