$$\lim_{x \rightarrow + \infty} ( \frac{ x }{ x-1 } - \frac{ 1 }{ \ln (x) })$$

Question

shubhamsrg · Answer

at x-> inf, 1/ln(x) is ofcorse 0

so your main problem must be with x/(x-1) right?

anonymous · Answer

yes

parthkohli · Answer

L'Hopital baby!

shubhamsrg · Answer

You must be knowing LH rule ?

shubhamsrg · Answer

lol..

anonymous · Answer

yes i know

shubhamsrg · Answer

So, where are you stuck then ?

parthkohli · Answer

$$\dfrac{d}{dx} x = \cdots$$and$$\dfrac{d}{dx} x - 1 = \cdots$$Though I first realized the answer through a good trick.

anonymous · Answer

i want answer only

shubhamsrg · Answer

If you want the answer,then get your answer! ;)

parthkohli · Answer

Basically, if you look at the fraction $\dfrac{x}{x  -1 }$, you will realize that when $x$ grows, then $\dfrac{x}{x - 1}$ grows too. But it never really reaches $1$, but it does keep approaching it!

shubhamsrg · Answer

you may even treat x/(x-1) as 
(x-1)+1 /(x-1) = 1 + 1/(x-1)

now put x->inf.

parthkohli · Answer

Like $x = 2 \iff \dfrac{x}{x - 1} = 2$.
$$x = 1000 \iff \dfrac{x}{x - 1} = \dfrac{1000}{1001} $$It's getting closer and closer to $1$.

shubhamsrg · Answer

Well thats the whole importance of limits @ParthKohli rightly says here.

Suppose we have
x-2 = x + 9 and we want to solve for x,

Many will simply say no solution; but if you talk about limits, x=inf is one answer.
Limits is all theoritical. 
Because of limits only we can say that 2 parallel lines meet at infinity or vice versa.

shubhamsrg · Answer

Your case is similiar
x and x-1 are parallel lines.
but x=x-1at x->inf

hence the ratio =1 at inf

parthkohli · Answer

Yes, I like theoretical stuff 8-)