# How do you find the limit #(u-1)^3/((1/u-u^2+3u-3)# as #u->1#?

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To find the limit of the expression (u-1)^3/((1/u-u^2+3u-3) as u approaches 1, we can simplify the expression and then substitute the value of u into it.

First, let's simplify the expression: (u-1)^3 = (u-1)(u-1)(u-1) = (u^2-2u+1)(u-1) = u^3-3u^2+3u-1

(1/u-u^2+3u-3) = (1/u) - u^2 + 3u - 3

Now, substitute u=1 into the simplified expression: (1^3-3(1)^2+3(1)-1) / (1/1-1^2+3(1)-3) = (1-3+3-1) / (1-1+3-3) = 0/0

Since we obtained an indeterminate form of 0/0, we need to further simplify the expression to evaluate the limit. To do this, we can factorize the numerator and denominator:

Numerator: u^3-3u^2+3u-1 = (u-1)(u^2-2u+1) = (u-1)(u-1)^2 = (u-1)^3

Denominator: (1/u) - u^2 + 3u - 3 = (1/u) - (u^2 - 3u + 3)

Now, substitute u=1 into the factored expression: (1-1)^3 / (1/1 - (1^2 - 3(1) + 3)) = 0 / (1 - (1 - 3 + 3)) = 0 / (1 - 1) = 0 / 0

Again, we obtained an indeterminate form of 0/0. This suggests that we need to further simplify the expression. To do this, we can use L'Hôpital's rule, which states that if we have an indeterminate form of 0/0 or ∞/∞, we can differentiate the numerator and denominator until we obtain a determinate form.

Differentiating the numerator and denominator: Numerator: d/dx[(u-1)^3] = 3(u-1)^2 * du/dx = 3(u-1)^2 Denominator: d/dx[(1/u) - (u^2 - 3u + 3)] = d/dx[1/u] - d/dx[u^2 - 3u + 3] = -1/u^2 - (2u - 3)

Now, substitute u=1 into the differentiated expression: 3(1-1)^2 / (-1/1^2 - (2(1) - 3)) = 0 / (-1 - (2 - 3)) = 0 / (-1 - 2 + 3) = 0 / 0

We still have an indeterminate form of 0/0. To further simplify, we can apply L'Hôpital's rule again by differentiating the numerator and denominator:

Differentiating the numerator and denominator: Numerator: d/dx[3(u-1)^2] = 6(u-1) * du/dx = 6(u-1) Denominator: d/dx[-1/u^2 - (2u - 3)] = d/dx[-1/u^2] - d/dx[2u - 3] = 2/u^3 - 2

Now, substitute u=1 into the differentiated expression: 6(1-1) / (2/1^3 - 2) = 0 / (2 - 2) = 0 / 0

We still have an indeterminate form of 0/0. To further simplify, we can apply L'Hôpital's rule for the third time by differentiating the numerator and denominator:

Differentiating the numerator and denominator: Numerator: d/dx[6(u-1)] = 6 Denominator: d/dx[2/u^3 - 2] = d/dx[2/u^3] - d/dx[2] = -6/u^4

Now, substitute u=1 into the differentiated expression: 6 / (-6/1^4) = 6 / (-6) = -1

Therefore, the limit of the expression (u-1)^3/((1/u-u^2+3u-3) as u 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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