In this video , I showed how to evaluate the limit of a rational exponential function. In this video, I highlighted the need to develop good algebra skills before attempting Lhopital's rule
I wish I’d had such a maths teacher, you’re so considerate and benevolent. Having pupils feel loved and cared for is the first essential element in pedagogy. Not for the sake of kindness alone but because of what that means about teacher’s ability to understand the needs of pupils. The other element of course is competence in the subject matter. You are endowed with both.
Amazing video, as always!! Just a last-minute idea: I believe that another interesting (though messy) way to solve these, is through Taylor series... in theory, "a^x = e^(x ln a)". Perhaps, in the end, you have 3 power series above and below, and you could join them. All of them have the same max-degree term approaching infinity at the same pace, so you may end up with a typical infinity/infinity case which when approached symbolically, may end up with the same right result!
But it's symmetrical... so you could just go ahead and rewrite the fraction so the top and bottom lines up. lim x-> -inf. ( (9^x - 8^x + 7^x) / (9^x + 8^x - 7^x) ) And then you cancel like terms by simply dividing leaving: lim x->inf. ( 1 - 1 - 1) which is just 1 - 2 = -1 way easier and quicker without much thaught.
You can't cancel terms like that. Suppose we have (a - b + c)/(a + b - c), by what you just did, we get 1 - 1 - 1 = -1. But if I put in a = 2, b = 6, c = 10, we get 6/-4 = -1.5, but by your logic, we have -1.
You could have done all the simplifying first and then did the change of variable it necessary. I also saw that everything was in terms of a^x, such that taking the ln of the terms to pull out x to the front would have been an easier way to approach this.
Im proud of myself😊. I saw Lhopitals rule wouldnt work. So i tried the ration function approach. My only mistake was I multiplied by 1/9^x (I didnt transition to t) instead of 1/7^x. But thats an easy mistake to fix
Me who forgets about L'Hopital everytime, this time: 🦍🦍🦍 Another reason that L'Hopital won't work according to me is that n^x's derivative returns n^x*ln(n) so, the derivative function is technically nearly similar
LOL, I tend to forget STEP ONE, which is to apply direct substitution. I've lost count of the times I found myself lost in a jungle of algebra, just to realize that all I needed to do was "plug it in". They say we learn from our mistakes. Well, I guess this is my personal exception to that rule :)
Sneaky! I exhausted pretty much every algebraic trick in the book, and even tried L'Hôpital's rule as a second-to-last resort, before realizing that all I had to do was think about dominant terms. Even then, I thought 9^𝑥 was the dominant term, so I divided everything by 9^𝑥. But about halfway through, I realized that since 𝑥 is approaching _negative_ infinity it's actually 7^𝑥 that is the dominant term. And sure enough, dividing everything by 7^𝑥, the problem basically solved itself. Oof.
