Limx tends to 1 (1 / lnx-x / LNX)

Limx tends to 1 (1 / lnx-x / LNX)

Limx tends to 1 (1 / lnx-x / LNX)
= lim (1-x)/lnx
=lim (-1)/(1/x) = lim(-x) = -1

Find limx → 1 [x / (x-1) - 1 / LNX]
Such as the title

Then, by using the lobida's law Lim {(x / x-1) - (1 / LNX)} = Lim [(xlnx-x + 1) / (x-1) LNX] to obtain the molecular denominator of LIM [(LNX + 1-1) / (LNX + 1-1 / x)] = Lim [(LNX) / (LNX + 1-1 / x)], Lim [(1 / x) / (1 / x + x ^ (- 2))] is obtained

Find the approximate solution of the equation LNX + x-3 = 0 in (2,3). (accurate to 0.1)

Let f (x) = LNX + x-3, that is, find the zero point of function f (x) in (2,3). Because f (2) = ln2-1 ＜ 0, f (3) = Ln3 ＞ 0, that is, (2,3) as the initial interval, the dichotomy list is as follows: degree left endpoint left endpoint function value right endpoint right endpoint function value interval length the first time 2-0.30685 31.09861 1 1 the second time 2-0.30685 2.50.41629 0.5 the third time 2-0.30685 2.25 0.06093 0.25 the fourth time 2.125-0.12123 2.25 0.06093 0.125 the fifth time 2.1875-0.02974 2.25 0.06093 0.0625 the sixth time 2.1875-0.02974 2.21875 0.01569 0.03125 because all the values in the interval (2.1875, 2.21875) are accurate to 0.1 and are 2.2, the approximate solution of the equation is 2.2

Given that the root of equation LNX + X-2 = 0 is x0, then the root of equation LNX + EX-1 = 0 is x0

X0 is the root of the first equation, lnx0 + x0-2 = 0
Let t = x0 / E and substitute it into the left side of the second equation
lnt+et-1=lnx0-1+x0-1=lnx0+x0-2=0
So the root of the second equation is: x0 / E