# How do you find the limit of #((2x^2-6)/(5x-x^2))# as x approaches infinity?

This is one way to approach this problem :

Note: As the denominator get larger, the number will be smaller, and almost close to 0. We can stated as follow

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The other method is to use L'Hopitals' Rule

If we direct substitute we will get an intermediate form

We can differentiate again to get

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To find the limit of ((2x^2-6)/(5x-x^2)) as x approaches infinity, we can divide both the numerator and denominator by x^2. This gives us (2-6/x^2)/(5/x-1). As x approaches infinity, 6/x^2 approaches 0 and 5/x approaches 0. Therefore, the limit simplifies to 2/0, which is undefined.

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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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- How do you find the limit of #[sqrt(x+1) - 1]/[x]# as x approaches 0?

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