How do you find the limit of #(x^2+3)/(x^2+4)# as #x->-oo#?
graph{[-10, 10, -2, 2]} graph{(x^2+3)/(x^2+4)
which agrees with the graph above perfectly.
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In light of L'Hospital's Rule,
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To find the limit of (x^2+3)/(x^2+4) as x approaches negative infinity, we can divide both the numerator and denominator by x^2. This gives us the limit of (1 + 3/x^2)/(1 + 4/x^2) as x approaches negative infinity. As x approaches negative infinity, both 3/x^2 and 4/x^2 approach zero. Therefore, the limit simplifies to 1/1, which is equal to 1. Hence, the limit of (x^2+3)/(x^2+4) as x approaches negative 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.
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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