1) Derive the numerator and denominator respectively by using the Lobida rule =(2X-7)/(2X-5),

So x→4, the original formula → 1/3.

2) Comprehensive score = (x+2) (1-x)/(x+1) (x 2+x+1) =-(x+2)/(x 2+x+1.

So x→ 1, the original formula →-1.

3) The total score is = (e x- 1-x)/(xe x-x), and the second derivative of the numerator denominator is obtained by the Lobida rule = 1/(x+2) respectively.

So x→0, the original formula → 1/2.

4) The original formula is = x 3/(x 2+2x+1), and the denominator of the molecule is deduced twice by Robida's law to get =3x respectively.

So x→∞, the original formula →∞

5) Using L'H?pital's law, the denominator of the molecule is deduced three times = (e x+e (-x))/cosx.

So x→0, the original formula →2.

6) When x→a and a-x→0, so sin (1/a-x) →1/a-x.

So x→a, the original formula →-1

7) When x→0, sin4x→4x, In( 1+2x)→2x, the original formula →2.

So x→0, the original formula → 2.

8) The original formula = (x 2+8x+ 16)/x, and the numerator and denominator are respectively deduced by the Lobida rule to obtain =2x+8.

So x→∞, the original formula →∞

9) When x→0, arcsinx→x, so the original formula =3/5.

So x→0, the original formula →3/5.

10) The second derivatives of numerator and denominator are obtained by using the Lobida rule = 1 respectively.

So n→∞, the original formula→1

11) f (0)' = (f (5t)-f (-t))/6t =1,and the original formula =3f(0)'=3.

12) Because f(x) exists and is finite, x= 1 is a root of x 3-x 2-ax+4, that is,1-kloc-0/-a+4 = 0, and a=4 is obtained.

I hope I can help you.