In your last example how exactly did you factor 2x^3 + 3x + 1 to get (2x +1)(x + 1)? If I'm not mistaken you cannot factor this. Perhaps you meant 2x^2 and not 2x^3? Other than that good vid. Thank you for posting.
A slight correction for the first technique. When you factor out the numerator and cancel one of the factors with the denominator, you are not changing the function. The domain of a function is a part of its definition. When you cancel out that factor, the technically correct thing to write would be lim x->0 (x+3), x ≠ 2.
Hello Sir, this is really helpful but there is a little mistake at the last exercise the power of 2x should be 2 not 3, if it's 3 the function Does Not Exist. Thank you!
Multiply the denominators together and you get t^2(t+1). Since cross multiplying the fractions gives a t variable in each term in the numerator, the t^2 is reduced to t because t/t^2 = 1/t.
lim as x approaches -1 (2x^3+3x+1)/(x^2-2x-3) is wrong. Not equal to 1/4. limit does not exist. If it read (2x^2..............) lim as x approaches -1 would be 1/4
You say in the first example that the curves of x^2+x-6/x-2 and x+3 are not the same and you even stress this point is important. This shook my very mathematical core and I spent longer than I would like to admit trying to work out why the curve had changed but it hasn't. The whole point is that y=x^2+x-6/x-2 is equivalent to y=x+3. The only difference is that y=x+3 is continuous and therefore we can find a value at 2.
Around the time 2:45 you say graph for f(x) = (x² + x - 6) / (x - 2) and x+3 is completely different. This statement is not true. In fact it is almost the opposite, almost completely the same.Except at x=2 they are same. This type of presentation mistake mislead the students, because the essence of taking limit algebraically based on equivalency of the functions.
I think he should have used another way to describe the difference. In fact the functions are different because of that point x = 2 where the original function is not defined. By using the factoring method you come up with a related function where that point is defined and then you are able to calculate the limit, which is the same for x+3 and (x² + x - 6) / (x - 2). The same situation happens with the next example. ((x^4)^2-16/x) is not defined at x = 0 but x - 8 is defined there. The limits are the same though.
By the way, terrific work TrevTutor. Your courses are excellent. I'm brushing up on some topics and looking forward to starting your discrete mathematics course.
Good video. Because it is a only channel explain in English ( ch- limits and derivatives) Please do the video more and more. It is really helpful video 🙏
the only question I have is, is there a point where we decide that its limit doesn't exist, or do we keep simplifying till we have reached the simplest form?
