# How do you determine the limit of #sqrt(x-4)/(3x+5)# as x approaches negative infinity?

Notice that you have a square root in your numerator. This can complicate the usual process, so you'll need to take a few extra steps to make sure this works.

For the numerator, we have to make one small simplification to get the right number into the square root:

Now, we divide through and simplify:

Now, you just go ahead and take your limit:

Hope that helped :)

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To determine the limit of sqrt(x-4)/(3x+5) as x approaches negative infinity, we can divide both the numerator and denominator by x. This gives us sqrt(1-4/x)/(3+5/x). As x approaches negative infinity, 4/x approaches 0 and 5/x approaches 0. Therefore, the limit simplifies to sqrt(1-0)/(3+0), which is equal to sqrt(1)/3. Thus, the limit of sqrt(x-4)/(3x+5) as x approaches negative infinity is 1/3.

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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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