First of all, when lim tends to 0, the upper and lower parts are both 0, which is the 0/0 type of L'H?pital's law. Therefore, up and down at the same time; It becomes e x- 1/sinx+xcosx. At this point, x tends to zero, which is still 0. Therefore, the law of Lobida is implemented again. Deduce again. Change it to e x/cosx+cosx-xsinx. Here we go. When x approaches 0, this formula is equal to 1/2. The answer is 1/2.
the second question
The method of this problem, like the first one, is 0/0, and L'H?pital's law is used many times. Deduce semicolons up and down many times.
The first derivative is equal to sec2 (x)-11-cosx. After deduction, the semicolon is still 0 up and down.
The second derivative is equal to 2 seconds 2 (x) * tanx/sinx. After this deduction, the semicolon is still 0 up and down.
The derivative is also equal to 2/cos 2 (x) = 2.
The answer to this question is equal to 2.
Substitution formula of equivalent infinitesimal;
When x approaches 0:
e^x- 1 ~ x
ln(x+ 1) ~ x
sinx ~ x
arcsinx ~ x
tanx ~ x
arctanx ~ x
1-cosx ~ (x^2)/2
tanx-sinx ~ (x^3)/2
( 1+bx)^a- 1 ~ abx
L'H?pital's law:
Lobida's law is a method to determine the value of indefinite formula by deducing the numerator and denominator respectively and then finding the limit under certain conditions. As we all know, the limit of the ratio of two infinitesimals or the ratio of two infinitesimals may or may not exist.
1. The numerator and denominator tend to zero or infinity. Second, whether the numerator and denominator are differentiable within the definition domain.
When both conditions are satisfied, take derivative and judge whether the limit after derivative exists: if it exists, get the answer directly; If it doesn't exist, it means that this indefinite form can't be solved by L'H?pital's law. If it is uncertain, that is, the result is still undecided, then continue to use the Lobida rule on the basis of verification.