At the of 75 in a retired life I am enjoying your brilliant way of teaching mathematics; best wishes for your great work (from a senior citizen of India)
My brother you have explained the concept of squeeze theorem so simple thank so much pliz continue to make more videos you are a great tutor I have ever watched.
Apart from the clear way of your explanations you have a soothing voice which is just perfect. I am very glad I have found your videos. I taught math for many years.
If it were not for the mandatory use of squeeze theorem, could you have said that sin(5/x) would go to an undetermined value between -1 and 1, which, when multiplied by x⁶ which would go to 0, would also become 0
It doesn't matter because if negative x^5 will be negative and so will change the inequality signs on both sides but that would then retain the original inequality just written from right to left and you can rewrite it. So, the short answer is that nothing will happen.
I dont understand only one part though: How did you multiply the entire inequality with x^6 because thats basically 0? Or its not? Maybe its allowed because its very close to 0 from the right (positive) side? Please clarify on this.
The first example is quite common. Multiply x by 8 then multiply sin(8x) by 8. So the limit is 8. For the second, do the same think using sqrt t instead of 8. You should get 0 answer.
That first one requires a whole ‘nother one or two videos to explain. That second may require some dlc we learned in calc two…or you would at least have to come back to it after you learn derivatives and interpret it as one.
Ok for that second one you can do it with raw limits my argument is you can say sin x0+ sin x/sqrt(x) 0 x/sqrt x because your dividing more cookies by the same number of people Now we need a lower bound argument for the squeeze theorem. The limit is positive because positive/positive is positive for my lower bound I can just use the opposite lim x->0+ -x/sqrt(x) < L < lim x->0 x/sqrt(x). Since x/sqrt(x)=sqrt(x) -sqrt(0)0- sin(x)/sqrt(x) lim u->0+ sin(-u)/sqrt(-u) lim u->0+ -sin(u)/(i*sqrt(x)) Lim u->0+ (-1/i)*[sin(u)/sqrt(u)] Lim u-> 0+ (-1/i)*[0]= 0
