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

Gather terms into a single ratio and apply L'Hopital's rule twice to find

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To find the limit of the given expression as x approaches 1, we can use algebraic manipulation and the properties of limits.

First, let's simplify the expression:

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

To combine the fractions, we need a common denominator. The common denominator is (x-1)ln(x).

Multiplying the first fraction by ln(x)/ln(x) and the second fraction by (x-1)/(x-1), we get:

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

Now, we can combine the fractions:

[(xln(x)) - (x-1)] / [(x-1)ln(x)]

Next, we can simplify further:

[xln(x) - x + 1] / [(x-1)ln(x)]

Now, we can factor out x from the numerator:

x(ln(x) - 1 + 1/x) / [(x-1)ln(x)]

Canceling out the common factor of (x-1) in the denominator and numerator, we have:

x(ln(x) - 1 + 1/x) / [ln(x)]

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

Substituting x = 1 into the expression, we get:

1(ln(1) - 1 + 1/1) / [ln(1)]

Since ln(1) = 0, the expression becomes:

1(0 - 1 + 1/1) / [0]

Simplifying further, we have:

1(-1 + 1) / [0]

Which simplifies to:

0 / [0]

This is an indeterminate form, so we need to apply L'Hôpital's rule.

Differentiating the numerator and denominator with respect to x, we get:

d/dx [x(ln(x) - 1 + 1/x)] / d/dx [ln(x)]

Using the product rule and chain rule, we have:

[(ln(x) - 1 + 1/x) + x(1/x - 1/x^2)] / (1/x)

Simplifying further, we get:

[(ln(x) - 1 + 1/x) + (1 - 1/x)] / (1/x)

Combining like terms, we have:

[ln(x) - 1 + 1/x + 1 - 1/x] / (1/x)

Simplifying, we get:

[ln(x)] / (1/x)

Multiplying by x/x, we have:

[xln(x)] / 1

Which simplifies to:

xln(x)

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

Substituting x = 1 into the expression, we get:

1ln(1)

Since ln(1) = 0, the expression becomes:

1(0)

Which equals:

0

Therefore, the limit of x/(x-1) - 1/(ln(x)) as x approaches 1 is 0.

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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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