How do you find the limit of #(x^2+x-6)/(x^2-9)# as #x->-3#?
Factorize the numerator and denominator:
so:
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To find the limit of (x^2+x-6)/(x^2-9) as x approaches -3, we can substitute -3 into the expression and simplify. By substituting -3 for x, we get (-3^2+(-3)-6)/((-3)^2-9). Simplifying further, we have (9-3-6)/(9-9), which becomes 0/0. This is an indeterminate form. To evaluate the limit, we can factorize the denominator as (x+3)(x-3) and simplify the expression to (x-2)/(x+3). By substituting -3 into this simplified expression, we get (-3-2)/(-3+3), which simplifies to -5/0. This is still an indeterminate form. To resolve this, we can use L'Hôpital's rule by taking the derivative of the numerator and denominator separately. The derivative of (x-2) is 1, and the derivative of (x+3) is also 1. By substituting -3 into the derivatives, we get 1/1, which equals 1. Therefore, the limit of (x^2+x-6)/(x^2-9) as x approaches -3 is equal to 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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