# How do you evaluate the integral #int 1/x dx# from 0 to 1?

you can't because it does not converge. try doing it.

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To evaluate the integral ∫(1/x) dx from 0 to 1, we integrate the function 1/x with respect to x from the lower limit 0 to the upper limit 1. The integral of 1/x is ln|x|, where ln denotes the natural logarithm. Evaluating this integral from 0 to 1, we get ln(1) - ln(0). Since the natural logarithm is undefined at x = 0, we cannot evaluate the integral directly at that point. However, as x approaches 0 from the positive side, ln(x) approaches negative infinity. Therefore, ln(0) is undefined. Thus, the integral evaluates to ln(1) - (−∞) = 0 - (−∞) = ∞. Therefore, the integral of 1/x from 0 to 1 is divergent.

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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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