You sir are not only an engaging and insightful teacher, but also incredibly talented at free handing graphs! Looking forward to future posts!! Thank you sincerely 😊
At 9:03 you say that the rate that (cos(h) -1) approaches zero is faster than h. L'Hopital would disagree with your statement. Happy to be corrected as I tussled with what you said (I am no expert, so my argument is not from authority). It's a tortoise (cos(h)-1) and the (h)are scenario. To meet at zero at the same time the tortoise needs to be much (infinitesimally) closer to the finish line because she is approaching so much slower, as it were, than the hare! Good on you for sharing your teaching, hopefully I have expressed my thoughts clearly enough for you to come back and correct my thinking.
stephen castleden, L'Hopital certainly does not disagree. Applying L'Hopital to (cos h - 1) / h --> - sin h / 1. Clearly you see that sin h approaches zero more quickly than does 1.
They both approches zero but they do not have the same rate(orspeed) look at the of cosh-1/h and of sinh then you will understand that the cosh-1 approches zero(reaches the x axis faster than h) but in another case the h and sinh approches zero(reaches axis) simultaneously(at a same time)
Nole Cuber because in that case h is only approaching 0 not actually 0 therefore multiplied by sinx will approach sinx so he can dismiss it and simply accept that it is sinx.
You sir are not only an engaging and insightful teacher, but also incredibly talented at free handing graphs! Looking forward to future posts!! Thank you sincerely 😊
I enjoy watching this guy!
At 9:03 you say that the rate that (cos(h) -1) approaches zero is faster than h. L'Hopital would disagree with your statement. Happy to be corrected as I tussled with what you said (I am no expert, so my argument is not from authority). It's a tortoise (cos(h)-1) and the (h)are scenario. To meet at zero at the same time the tortoise needs to be much (infinitesimally) closer to the finish line because she is approaching so much slower, as it were, than the hare!
Good on you for sharing your teaching, hopefully I have expressed my thoughts clearly enough for you to come back and correct my thinking.
stephen castleden, L'Hopital certainly does not disagree.
Applying L'Hopital to (cos h - 1) / h --> - sin h / 1.
Clearly you see that sin h approaches zero more quickly than does 1.
That was π Day....!
14th march
What is the rsn behind writing cos and sin separately
Super
Aww, no Squeeze Theorem?
11:42
7:27 Did not understand how both are approaching and not approaching zero...
They both approches zero but they do not have the same rate(orspeed) look at the of cosh-1/h and of sinh then you will understand that the cosh-1 approches zero(reaches the x axis faster than h) but in another case the h and sinh approches zero(reaches axis) simultaneously(at a same time)
I hope it would help and sorry for my bad english
I just don't understand how in cosh-1/h isn't undefined because you plugged in 0 for h
Nole Cuber because in that case h is only approaching 0 not actually 0 therefore multiplied by sinx will approach sinx so he can dismiss it and simply accept that it is sinx.
Some time these ohhhh in bgs are totally trigger us what she has understood and we did't..