Question

Simple Question on LIMITS

Answer 1

\[\huge \bf \lim_{x \rightarrow \infty} \frac{x~\ln(1+\frac{lnx}{x})}{lnx}\]

Answer 2

i guess 1
can we use l'hopital or do we have to use some arcane method?

Answer 3

no ! don't use l-hospital rule.
i don't know the answer !

Answer 4

can you simplify this by using SPECIAL LIMITS ?

Answer 5

why all the "no l'hopital's rule" restrictions? seems like a perfectly good method to me!

Answer 6

i hate that rule !
its the easiest way to approach this and i don't like to approach by using easiest way because i have to fight biggest exam in the country !

Answer 7

HINT to this question :-
USE SPECIAL LIMITS i.e
\[\large \bf \lim_{x \rightarrow \infty}\frac{lnx}{x}=0\]

Answer 8

enjoy!

Answer 9

lol

Answer 10

\[\frac ab \implies \frac{1}{b/a}\]

Answer 11

\[\huge \bf \lim_{x \rightarrow \infty}\frac{\ln(1+\frac{lnx}{x})}{\frac{lnx}{x}}\]

Answer 12

then

Answer 13

now the limit shows it goes to 0/0, so your hint is actually hinting at Lhop ... use it

Answer 14

how the limit tends to 0/0 ?

Answer 15

what does your hint say?

Answer 16

as x to inf, lnx/x to 0

let a = lnx/x; as x to inf, a to 0

ln(1+a) ln(1)
------ ==> ------
a 0

Answer 17

ok. so its 0/0 form

Answer 18

the limit is 1 (numerically)

Answer 19

how ?

Answer 20

lhop gives us

a'
-------
a'(1+a)


1/(1+lnx/x) to 1 if i see it right

Answer 21

@amistre64 don't use l-hopital rule !

Answer 22

\[\begin{align} \lim_{x \rightarrow \infty} \frac{x~\ln(1+\frac{\ln x}{x})}{\ln x}
&= \lim_{x \rightarrow \infty} \ln\left(1+\frac{\ln x}{x}\right)^{1/(\ln x/x)}\\~\\
&=\ln~\lim_{x \rightarrow \infty} \left(1+\frac{\ln x}{x}\right)^{1/(\ln x/x)} \\~\\
&=\ln~\lim_{t \rightarrow 0} \left(1+t\right)^{1/t} \\~\\
&=\ln~e\\~\\
&=1
\end{align}\]

Answer 23

the hint suggests using lhop, if you want to sit there on your big exam and waste all your time on something else ..... its your choice

Answer 24

Let u = ln(x) /x , as x->oo , u -> 0

Answer 25

lol @amistre64
in that exam, there is not a single question that can solve by L-HOPITAL'S RULE !

that's why, i don't use it !

Answer 26

so thats your reason, ok

Answer 27

yep

Answer 28

then why are you practicing with this one?

Answer 29

it is a question without using l-hopital's rule ?

Answer 30

big number appraoch from perl is valid to me, and rational seems to have some secret knowledge as well.

Answer 31

$$
\Large \rm {
\lim_{x \rightarrow \infty}\frac{\ln(1+\frac{lnx}{x})}{\frac{lnx}{x}}

\\~\\ Let u = \frac{\ln x }{x} , ~ as ~x \to \infty , u \to 0
\\~\\
\lim_{u \rightarrow 0}\frac{\ln(1+u)}{u}
}
$$

Answer 32

so it form 0/0 form

Answer 33

then, next @perl

Answer 34

that is a given limit, ln(1 + x ) / x -> 1 as x->0

Answer 35

i dont have any secrt knowledge lol
the limit definition of "e" is known to almost everyone.. including statisticians and politicians ;p

Answer 36

\[\large \bf \frac{\ln(1+u)}{u}=\frac{0}{0}form\]

Answer 37

use lhop if your exam is mcq type

Answer 38

cant we do power series for ln(u+1) and divide off a u?

Answer 39

@rational IIT JEE is that exam

Answer 40

if it is a big boy's exam, then they should allow u to use lhop and taylor series

Answer 41

$$\large \ln x' \vert_{x=a=1} = \lim_{h \rightarrow 0} \ \frac{\ln(1+h) - \ln 1}{h} \ .$$

Answer 42

@amistre64 i stuck !