# Does the function converge or diverge #1/((x-1)(x^2+1)) #on the bound [2,infinity]?

Assuming poles are meant,

The nomenclature is a bit unclear - the given function doesn't really "converge" or "diverge" in the way it is specified - these are more properties of sequences or series. Functions can be specified as series, a manner in which they can converge or diverge for various values, but that doesn't seem to apply here.

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To determine if the function ( \frac{1}{(x-1)(x^2+1)} ) converges or diverges on the interval ([2, \infty)), we need to check for convergence or divergence of the function as ( x ) approaches infinity. We can do this by analyzing the behavior of the function at large values of ( x ).

As ( x ) approaches infinity, the terms ( (x-1) ) and ( (x^2+1) ) become dominant. Since ( x^2 ) grows faster than ( x ), the term ( (x^2 + 1) ) dominates the denominator.

Therefore, as ( x ) approaches infinity, the function behaves like ( \frac{1}{x^2} ), which converges. Hence, the function ( \frac{1}{(x-1)(x^2+1)} ) converges on the interval ([2, \infty)).

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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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