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

Answer 1

Charles Anctil

#1/3#

#sin(x-1)/(x^2+x-2)= sin(x-1)/((x+2)(x-1)) = (sin(x-1)/(x-1)) 1/(x+2)#

#lim_(x->1)sin(x-1)/(x^2+x-2)=lim_(x->1)(sin(x-1)/(x-1)) 1/(x+2) = 1 cdot 1/3#

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Answer 2

Isaiah Allen

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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Answer from HIX Tutor

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.

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

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