LIM (1 / ln (1 + x) - 1 / x) x tends to 0

LIM (1 / ln (1 + x) - 1 / x) x tends to 0

lim(1/ln(1+x)-1/x)
=lim[x-ln(1+x)]/xln(1+x)
=lim[1-1/(1-x)]/[ln(1+x)+x/(1+x)]
=limx/[x+(1+x)ln(1+x)]
=lim1/[ln(1+x)+1+1]=1/2

How to use lobita's law to find: LIM (x - > 1 +) (LN x) * ln (x-1)
If I change 1 / ln x into denominator, I can get the limit 0, but why can't I change 1 / ln (x-1) into denominator? In addition, can I use the combination of equivalent infinitesimal to do it

lim(x->1+) (ln x) * ln(x-1)
=lim(x->1+) (ln x) / [1/ln(x-1)]
Obviously, when X - > 1 +, LN X - > 0, and ln (x-1) tends to - ∞, so [1 / ln (x-1)] tends to 0,
By deriving the numerator and denominator at the same time, we get
Original limit
=lim(x->1+) (1/x) / [-1/ (x-1)*ln²(x-1)]
=The denominator of LIM (x - > 1 +) ln & # 178; (x-1) / [1 / (x-1)] tends to infinity, and the derivative is obtained
=lim(x->1+) [2ln(x-1) /(x-1)] / [-1/(x-1)²]
=The denominator of LIM (x - > 1 +) - 2ln (x-1) / [1 / (x-1)] tends to infinity. At the same time, the derivative is obtained
=lim(x->1+) [-2/(x-1)] / [-1/(x-1)²]
=lim(x->1+) 2(x-1)
=0
This is obviously a bit of a hassle
It's a lot easier to do it with an equivalent infinitesimal,
lim(x->1+) (ln x) * ln(x-1)
=lim(x->1+) ln(1+x-1) * ln(x-1)
Obviously, when X - > 1 +, X-1 - > 0 +,
So ln (1 + x-1) and X-1 are equivalent infinitesimals
therefore
Original limit
=lim(x->1+) (x-1) * ln(x-1)
=The denominator of LIM (x - > 1 +) ln (x-1) / [1 / (x-1)] tends to infinity. At the same time, the derivative is obtained
=lim(x->1+) 1/(x-1) / [-1/(x-1)² ]
=lim(x->1+) -(x-1)
=0

LIM (1 / ln (x + 1) - 1 / x) x tends to zero using the law of lobita
LIM (1 / ln (x + 1) - 1 / x) x tends to zero using the law of lobita

Original = [x-ln (x + 1)] / XLN (x + 1)
=[x-ln (x + 1)] / x ^ 2 [ln (x + 1) and X are equivalent infinitesimals when x tends to zero]
=[1-1 / (x + 1)] / 2x [0 / 0 lobita rule]
=x/2x(x+1)
=1/2

Lim → 0 [∫ (upper limit x, lower limit 0) ln (1 + Sint) DT] / 1-cosx

Lim → 0 [∫ (upper limit x, lower limit 0) ln (1 + Sint) DT] / 1-cosx
It can be known from the upper and lower derivatives of the law of Robita (the derivation of the integral with variable upper limit for molecules)
= lim→0[ln(1+sinx)]/sinx
From the equivalent infinitesimal ln (1 + SiNx) = SiNx
= lim→0 (sinx)/sinx
=1

What number is added to the numerator and denominator of 1 / 13 to make it 60%?

1/13
60%=6/10 =18/30
Plus 17

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Let's start with an example
Half of a pencil
Half of them are gone
Conclusion: the singular and plural of "half of" followed by countable nouns can be used

Function f (x) = | (2 ^ x) - 1 |, if AF (b), then the following four formulas hold
A.a

When - ∞

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As shown in figure I is the heart of △ ABC, the extension line of AI intersects at point D, the circumscribed circle intersects at point E, (1) is be equal to IE? Why? (2) Ie is the proportion of AE and de

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