# How do you find the limit of #sin(x-1)/(x^2+x-2)# as #x->1#?

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To find the limit of sin(x-1)/(x^2+x-2) as x approaches 1, we can use direct substitution. Plugging in x=1 into the expression, we get sin(1-1)/(1^2+1-2), which simplifies to sin(0)/0. Since sin(0) is equal to 0, the expression becomes 0/0. This is an indeterminate form, so we need to apply L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get cos(x-1)/(2x+1). Evaluating this expression at x=1, we have cos(1-1)/(2(1)+1), which simplifies to cos(0)/3. Since cos(0) is equal to 1, the limit becomes 1/3. Therefore, the limit of sin(x-1)/(x^2+x-2) as x approaches 1 is 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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