2 questions to find the limit. Master, please help me

Question 1: Use variables to replace t=(2)-x, then when x 2, t 0, the original formula = lim [t*tan( 2-t)]=lim[t*cot(t)]=lim[(t sint)*cost]=cost=1 (because t tends to 0 when lim(t sint)=1).

Question 2: When using an equivalent infinitesimal substitution x 0 (1-cosx) is equivalent to (1 2) x 2, so.

x 0 lim(x 2) (1-cosx) is equal to x 2) [(1 2) x 2]=2 and vice versa.

Finding the derivative, Robida's Law.

1: de1

2: de2

In the first question, put [(2)-x] into the denominator and write it as 1 [(2)-x], infinity is infinite, and Robida's law gives the answer 1 3 times

In the second question, since the equivalent infinitesimal of 1-cosx is x 2 2, it can be equivalent to replace the answer 2 directly

The second question is 2 when x approaches o, sinx approaches x, i.e. sinx 2 approaches x 2

And 1-cosx=2sin2(x2) approaches 2 times x2 times x2 equals x2 2

So the result is 2

Question 1: Replace tanx with cot(2-x) so that it becomes (2-x) tan(2-x), so the result should be 1, because when x approaches o, x approaches tanx

In the second question, you write 1-cosx as sinx 2 2 by lim(sinx x)=1, or lim(x sinx)=1, (all the same, when x tends to 0), so the answer is 2

The first question is a case of multiplying infinity by infinitesimal and also changing accordingly. I don't have a pen at this time, it's inconvenient to change, but I think it should be pi 2-x divided by 2, pi 4 + x 2 appears, and then tanx is also written in the form of sin (x 2) like the second question, and the result can be out.

Infinity Infinity, 0 0 cases, solved by Lobida's rule, is to find the derivatives separately and then compare.

Question 1: tgx, sinx, cosx

sinx=1 is not considered.

The remaining (pi 2-x) cosx derivatives up and down 1 1 1 2nd question. Derivative up and down 2x sinx and derivative 2 1 2

Line 3 is wrong, directly =e 2

Note: x tends to infinity, not to 0!

One more step, and it will be clear that the spring will collapse!

For reference, please laugh at Na Meng's family.

2. Use the product of the sum difference, the product of the sum difference, and the equivalent infinitesimal substitution.

Original formula = lim(n-> sin(1 n 2) + sin(2 n 2)+sin(n/n^2)]*2sin(1/2n^2)/2sin(1/2n^2)

lim(n->∞2sin(1/2n^2)sin(1/n^2)+2sin(1/2n^2)sin(2/n^2)+.2sin(1/2n^2)sin(n/n^2)]/(2*1/2n^2)

lim(n->∞n^2

lim(n->∞n^2

lim(n->∞2sin[(n+1)/2n^2]sin(n/2n^2)*n^2

lim(n->∞2*[(n+1)/2n^2]*(1/2n)*n^2

lim(n->∞n^2+n)/2n^2

These two questions have the same meaning, (2) the denominator is multiplied by the root number (x-2) + the root number under the root number 2

The next question is the same.