Find lim x->1+ of [(1/(x-1))-(1/lnx)].

Here is my work...
=(lnx-(x-1)) / ((x-1)(lnx))

=(lnx-1) / (lnx+ (x+1)/x)

This becomes(1/x) / ((1/x)+(1/x^2))

which becomes 1/ (1/x^2)
This equals 1/2. I understand the answer has to be -1/2, but I am having trouble figuring out where that negative sign is coming from. What have I missed? Thank you in advance for your assistance.

=(lnx-(x-1)) / ((x-1)(lnx))

How did you get to the next line?

To find the limit of the given expression, let's go through the steps one by one.

Starting from your last step:
```
1 / (1/x^2)
```
When we simplify this expression, we can rewrite it as:
```
x^2 / 1
```
Now, let's go back to the beginning and see where a mistake might have been made.
We have:
```
[(1/(x-1)) - (1/lnx)]
= [(lnx - (x-1)) / ((x-1)(lnx))]
```
The numerator is correctly written as `ln(x) - (x-1)`. However, the denominator should be `((x-1) * lnx)` instead of `((x-1) * (lnx + (x+1)/x))`. So, we need to correct that part.
```
[(lnx - (x-1)) / ((x-1) * lnx)]
```

Now, let's continue simplifying:
```
= (lnx - x + 1) / ((x-1) * lnx)
```
Now, let's analyze what happens as `x` approaches `1` from the right `(x -> 1+)`.
When `x` approaches `1` from the right side, `ln(x)` approaches `ln(1+) = 0+`. Therefore, `ln(x)` is positive and very close to `0`.

Now, let's analyze the numerator:
```
lnx - x + 1
```
As `x` approaches `1`, `lnx` approaches `0+` but `x` approaches `1` itself. So, the numerator becomes:
```
0+ - 1 + 1 = 0
```

And the denominator:
```
(x - 1) * lnx
```
As `x` approaches `1`, `x - 1` approaches `0`, but `lnx` approaches `0+`. So, the denominator becomes:
```
0 * 0+ = 0
```
Since the denominator approaches `0` and the numerator approaches `0`, we have an indeterminate form of `0/0`.

To evaluate the limit in this case, we can apply L'Hopital's Rule, which states that if we have an indeterminate form of `0/0`, we can differentiate the numerator and the denominator and then evaluate the limit again.

Differentiating the numerator and the denominator:
Numerator:
```
d/dx (lnx - x + 1) = (1/x) - 1
```
Denominator:
```
d/dx ((x - 1) * lnx) = [(1) * lnx + (x - 1) * (1/x)] = lnx + (x - 1)/x
```
Now, let's evaluate the limit again:
```
lim x->1+ [(1/x) - 1] / [lnx + (x - 1)/x]
```
Substituting `x = 1` into the differentiated numerator and denominator:
```
[(1/1) - 1] / [ln(1) + (1 - 1)/1)]
= [1 - 1] / [0 + 0]
= 0 / 0
```
As we can see, after differentiating, we still end up with an indeterminate form of `0/0`. This suggests that we need to apply L'Hopital's Rule again.

Differentiating the numerator and the denominator once more:
Numerator:
```
d/dx [(1/x) - 1] = -1/x^2
```
Denominator:
```
d/dx [lnx + (x - 1)/x] = (1/x) + ((1 * x - (x - 1))/x^2)
= (1/x) + (1/x)
= 2/x
```
Now, let's evaluate the limit again:
```
lim x->1+ [-1/x^2] / [2/x]
```
Substituting `x = 1` into the differentiated numerator and denominator:
```
[-1/1^2] / [2/1]
= -1 / 2
= -1/2
```

Therefore, the limit of the given expression as x approaches 1 from the right `(x -> 1+)` is `-1/2`.