Anybody here take high school math in the 60s or 70s with Houghton-Mifflin's math series by Mary Dolciani et. al.?
I remember these books fondly and wish to purchase them: Algebra 1: Structure and Method (1967) ; Algebra 2 & Trigonometry; Math Analysis, and the pre-algebra 8th Grade text, the name of which I cannot remember.
Did you dig these books, or what?
Also, for math lovers, what other subjects do you love as much? Mine are physics and German.
@mysteriousnyc another thing to notice is the result of lim x->0 [(1-cosx/x)) = 0/2=0 and lim x->0 [x/(1-cosx)] = 2/0=undefined. So, we cannot use it as the lim x->0 (sinx/x)=1 and its reciprocal.
The "squeezing" method is also sometimes called the sandwich theorem.
harriette313 4 days ago
The last problem must be = 1 (not 0)
keerujames 2 months ago
really good lecture. Thanks
keithl28 3 months ago
very good lecture
50jegadeesan 5 months ago in playlist calculus
well guyz i jst gave the exams of matric class or 10th grade
n thanks alot to this gr8 main becz of this i have learn 7lectures of calculus
and i learned 2 years earlier becuz it is started frm Bsc in Pakistan :) :)
friend37167 10 months ago
thank you! God bless :) <3
seanlolsean 1 year ago
Anybody here take high school math in the 60s or 70s with Houghton-Mifflin's math series by Mary Dolciani et. al.?
I remember these books fondly and wish to purchase them: Algebra 1: Structure and Method (1967) ; Algebra 2 & Trigonometry; Math Analysis, and the pre-algebra 8th Grade text, the name of which I cannot remember.
Did you dig these books, or what?
Also, for math lovers, what other subjects do you love as much? Mine are physics and German.
Fersomling 1 year ago
Wonderful series! However after trying a number of calculations and graphing the last exercise, I keep getting
Lim (t^2/(1-cos(t)^2) = 1 as t goes to 0, instead of 0 as Prof. Delaware indicates.
deugle 2 years ago 2
@deugle yes he is wrong because 1-cos^2 (t) is is sin^2 (t) Which changes the problem to the type Lim t >> 0 of t / sin (t) * t / sin (t) which = 1
mysteriousnyc 2 years ago 4
@mysteriousnyc another thing to notice is the result of lim x->0 [(1-cosx/x)) = 0/2=0 and lim x->0 [x/(1-cosx)] = 2/0=undefined. So, we cannot use it as the lim x->0 (sinx/x)=1 and its reciprocal.
iqbalnaved 1 year ago