# How do you find the limit of # (x^2+2x-1)/(3+3x^2)# as x approaches infinity?

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To find the limit of (x^2+2x-1)/(3+3x^2) as x approaches infinity, we can divide both the numerator and denominator by the highest power of x, which is x^2. This gives us (1/x^2 + 2/x - 1/x^2)/(3/x^2 + 3). Simplifying further, we get (1 + 2/x - 1/x^2)/(3/x^2 + 3). As x approaches infinity, the terms 2/x and 1/x^2 become negligible, since they approach zero. Therefore, the limit simplifies to 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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- Find the limit as x approaches infinity of #y=ln( 2x )-ln(1+x)#?
- For what values of x, if any, does #f(x) = 1/((x+12)(x-1)) # have vertical asymptotes?

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