# How do you determine the limit of #(x)/sqrt(x^2-x)# as x approaches infinity?

So we get

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To determine the limit of (x)/sqrt(x^2-x) as x approaches infinity, we can simplify the expression by dividing both the numerator and denominator by x. This gives us 1/sqrt(1-1/x). As x approaches infinity, 1/x approaches 0, so the expression simplifies to 1/sqrt(1-0), which is equal to 1/1, or simply 1. Therefore, the limit of (x)/sqrt(x^2-x) as x approaches infinity 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.

- How do you evaluate the limit #(x^5-1)/(x-1)# as x approaches #1#?
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- How do you prove limit of #2/sqrt(x-4)=oo# as #x->4^+# using the precise definition of a limit?
- What is the limit of #(4x^2+1)/(x^2+2x-3)# as x approaches -2?
- What is the limit of # (sinx)(ln4x)# as x approaches 0?

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