May I ask why LIM (x ^ 4-3x + 2) / (x ^ 5-4x + 3) [x tends to be 1] is not calculated correctly by the law of lobita LIM (x ^ 4-3x + 2) / (x ^ 5-4x + 3) [x tends to 1] After the upper and lower derivatives, the value of the limit is 1, but the answer given in the book is 0. That is to say, the law of lobita is not applicable here The method used in the book is to match the numerator denominator of the limit into the form of product, and then the numerator denominator is dropped by about one (x-1), and then the limit is brought in to get a value of 0. I don't quite understand what's wrong with the lobita rule here It seems that the problem lies in whether x ^ 4-3x + 2 is differentiable when x = 1. If this is not differentiable, how to determine?

When x tends to 1, the original formula is 0 / 0 type, and the numerator and denominator are derivable on R, so it can be calculated by the law of Robida, which is indeed 1. The answer is right. Even if it is solved according to the method of the answer, it is 1 molecule x ^ 4-3x + 2 = x ^ 4-x-2x + 2 = x (x ^ 3-1) - 2 (x-1) = x (x-1) (x ^ 2 + X + 1) - 2 (x-1) = (x-1) (x ^ 3 + x ^ 2 + X-2)

Finding the limit of x [ln (X-2) - ln (x + 1)] when limx tends to infinity

lim(x→∞) x[ln(x-2)-ln(x+1)]=lim(x->∞) [ln(x-2)-ln(x+1)]/(1/x)=lim(x→∞) [1/(x-2)-1/(x+1)]/(-1/x^2)=lim(x->∞) -[(x+1-x+2)/[(x-2)(x+1)]x^2=lim(x->∞) -3x^2/(x^2-x-2)=-3

Using lobida's law to find limx → positive infinity ln (1 + x ^ 2) / ln (1 + x ^ 4)

lim(x→+∞) ln(1+x^2)/ln(1+x^4)=lim(x→+∞) (x^4+1)/(2x^4+2x²)=lim(x→+∞) (1+1/x^4)/(2+2/x²)=1/2

Given X & # 178 / X quartic power + X & # 178; + 1 = 1 / 4, find 5x quartic power - 15x & # 178; + 5 / 3x & # 178;

X & # 178 / X quartic power + X & # 178; + 1 = 1 / 4
4X & # 178; = x fourth power + X & # 178; + 1
x²-3x²+1=0
5x quartic power - 15x & # 178; + 5 / 3x & # 178;
=5 (x fourth power - 3x & # 178; + 1) / 3x & # 178;
=0
If you don't understand this question, you can ask,

If we know the fourth power of X & # 178 / x + X & # 178; + 1 = 1 / 4, then the fourth power of 5x - 15x & # 178; + 5 / 3x & # 178=

Hello
The fourth power of X & # 178 / x + X & # 178; + 1 = 1 / 4
The fourth power of X + X & # 178; + 1 = 4x & # 178;
The fourth power of X + 1 = 3x & # 178;
x²+1/x²=3
The fourth power of 5x - 15x & # 178; + 5 / 3x & # 178;
=5 (the fourth power of X - 3x & # 178; + 1) / 3x & # 178;
=5/3(x²-3+1/x²)
=5/3x(3-3)
=0
I don't understand. I can ask
If you have any help, please remember to adopt it. Thank you
I wish you progress in your study!

The derivative of function f (x) at t, f '(T) = 0, is the condition for f (x) to reach the extremum at t?

It is neither sufficient nor necessary
The derivative of cubic function y = X3 at 0 is 0, but it is not an extreme point; y = absolute value x is a minimum at 0, but it is not differentiable here

The derivative problem: is the derivative value of y = f (x) at one point 0 a necessary and insufficient condition for y = f (x) to take the extremum at this point?
How to explain that the function at the apex gets the extremum at the apex
If the function y = f (x) is differentiable at a certain point, it seems to be a necessary and insufficient condition.

I don't think it's right. Is it necessary
That's what you call the pinnacle
In this case, by the definition of extremum, it is indeed extremum
But obviously, the left and right derivatives here are not equal, so they are not differentiable
So it's not necessary

For a differentiable function, the difference sign of derivatives on both sides of a point is a sufficient and unnecessary condition for this point to be an extreme value. Why, for example?
It's not necessary. Let's take a function

For a straight line, it can be derived everywhere, any point is an extreme point, but the derivatives on both sides of this point are not the same sign; the derivatives on both sides can deduce that this point is an extreme point, so it is a sufficient and unnecessary condition

The second derivative of y = e ^ (3x)

y'=e^(3x)*（3x)'=3e^(3x)
So y '' = 3 * [e ^ (3x)] '= 3 * 3E ^ (3x) = 9E ^ (3x)