L'Hospital's rule problem?

Question

lim ( x/(x-1) - 1/ln(x) )
x->1

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secondly, can you take the limits separately?

I tried the problem by using common denominator ln(x)...and end up with 1/2.

But I'm still skeptical. Can anyone compare this with me?

alright some people are posting wierd answers....remember guys, you can only take the l'H rule if 0/0, and infinity/infinity....

also, the common denominator was (x-1)lnx when i solved it....lemme Heres work:

lim __x(lnx) - x +1_
x-> 1 (x-1)lnx

LH rule:
lim __(lnx)x______
x-> 1 (x-1) + (lnx)x

LH rule again:

lim __lnx + 1______
x-> 1 1 + lnx + 1

= 1/2


Anyone agree?

Answer 1

No, you cannot take the limits separately.

Your answer of 1/2 is correct.

lim (x/(x-1) - 1/lnx)
x->1

=lim [(x ln x - x + 1)]/[(x-1)(ln x)]
x->1

after applying L'Hospital's rule you get
lim [x*(1/x) + ln x -1] / [(x-1)(1/x) + ln x]
x->1

=lim (ln x)/[1 - (1/x) + ln x ]
x->1

=lim (x ln x )/ [x - 1 + x ln x]
x->1

... and so on... whatever, you are correct.

Answer 2

No you cannot take the limits separately.

1.) Change lnx to 1/x and you'll get 1/1/x
2.) This will leave you with 1/1*x/1 which equals x.
3.) Multiply (x)*(x-1) for common denominator, and you'll get
=lim x/(x-1) - x(x-1)
x->

4.) Cancel out the (x-1) and you'll get x minus x.
5.) x-x=0 so you must use L'Hospital Rule in order to find the limit because you cannot factor.

Use L'Hospital Rule and see if you can figure it out.
L'Hospital rule = Derivative of the Limit.

Answer 3

= 1+1/(x-1)-1/ln(x)
Let´see 1/(x-1)-1/ln(x) = (ln x)-x+1]/(x-1)ln x
The derivative of the numerator is (1/x)-1
and of the denom. is ln x+(x-1)/x
Now we have lim (1-x)(/xlnx+x-1)
again = lim -1/( 1+lnx+1) = -1/2 which summed to 1 gives =1/2
Separately both summands have lim infinity so you can´t

Answer 4

address to phjun8383 there
http://answers.yahoo.com/question/index;_ylt=AqDwWCqUK25vnLR7OYCDDk7sy6IX?qid=20070811133941AANdImX