# How do you find the limit of #sqrt(9x+x^2)/(x^4+7)# as x approaches #oo#?

Reqd. Lim.

We need to recall, here, that,

The inter-changeability of the limit & sqrt. fun is because of the continuity of the sqrt. fun.

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The limit exists, and it is zero.

Thus, the square root approaches one:

As for the parenthesis in the denominator, with similar claims we have

Thus, the global ratio behaves like

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To find the limit of sqrt(9x+x^2)/(x^4+7) as x approaches infinity, we can use the concept of limits.

First, we divide both the numerator and denominator by x^2 to simplify the expression:

sqrt((9/x) + 1)/(x^2/x^2 + 7/x^2)

Simplifying further, we get:

sqrt((9/x) + 1)/(1 + 7/x^2)

As x approaches infinity, 9/x and 7/x^2 both approach zero. Therefore, the expression simplifies to:

sqrt(0 + 1)/(1 + 0)

Which is equal to:

sqrt(1)/1

And the square root of 1 is equal to 1. Therefore, the limit of sqrt(9x+x^2)/(x^4+7) 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.

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