# How do you find the limit of #(1/(ln x) - 1/(x-1))# as x approaches 1?

This limit 'creates' the

So (in this case using the de L'Hôpital's Rule twice):

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To find the limit of (1/(ln x) - 1/(x-1)) as x approaches 1, we can use algebraic manipulation and L'Hôpital's rule.

First, let's simplify the expression by finding a common denominator:

(1/(ln x) - 1/(x-1)) = ((x-1)/(ln x(x-1))) - (ln x/(ln x(x-1)))

Next, combine the fractions:

((x-1) - ln x)/(ln x(x-1))

Now, we can apply L'Hôpital's rule by taking the derivative of the numerator and denominator separately:

Numerator: d/dx ((x-1) - ln x) = 1 - 1/x Denominator: d/dx (ln x(x-1)) = (1/x)(x-1) + ln x

Simplifying the derivatives:

Numerator: 1 - 1/x Denominator: (x-1)/x + ln x

Now, we can evaluate the limit as x approaches 1 by plugging in the value:

(1 - 1/1)/((1-1)/1 + ln 1) = 0/0

Since we still have an indeterminate form, we can apply L'Hôpital's rule again:

Numerator: d/dx (1 - 1/x) = 1/x^2 Denominator: d/dx ((x-1)/x + ln x) = (1/x) - (x-1)/x^2 + 1/x

Simplifying the derivatives:

Numerator: 1/x^2 Denominator: 1/x - (x-1)/x^2 + 1/x

Now, we can evaluate the limit as x approaches 1:

(1/1^2)/(1/1 - (1-1)/1^2 + 1/1) = 1/1 = 1

Therefore, the limit of (1/(ln x) - 1/(x-1)) as x approaches 1 is 1.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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