# How do you determine if the improper integral converges or diverges #int (1/(3x)-6) dx# from negative infinity to 0?

To evaluate a two-sided improper integral, split it into two integrals and express each as a limit like so:

Now compute the integrals and evaluate the limits:

Technically, we could stop here. If one integral diverges, the whole expression will diverge. But let's evaluate the second limit just for good measure:

And again we see the integral also diverges as it approaches 0.

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To determine if the improper integral converges or diverges, we need to evaluate the limit of the integral as the upper limit approaches 0 from the left (since the lower limit is negative infinity). We can express the integral as follows:

∫(from -∞ to 0) (1/(3x) - 6) dx

To integrate (1/(3x) - 6), we need to split it into two separate integrals:

∫(from -∞ to 0) (1/(3x)) dx - ∫(from -∞ to 0) 6 dx

The first integral, ∫(1/(3x)) dx, is an improper integral that diverges. This can be shown by evaluating the limit of the integral as x approaches 0 from the left:

lim(x→0-) ∫(from -∞ to x) (1/(3x)) dx = lim(x→0-) [ln|3x|] (from -∞ to x) = lim(x→0-) [ln|3x| - ln|3(-∞)|] = ∞

The second integral, ∫(6) dx, is a finite integral:

∫(from -∞ to 0) 6 dx = [6x] (from -∞ to 0) = 6(0) - 6(-∞) = 0 - (-∞) = ∞

Since one part of the integral diverges (∫(1/(3x)) dx) and the other part converges (∫(6) dx), the overall integral diverges.

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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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