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Factor the numerator and simplify.
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x^2 +x-6=(x+3)(x-2)

so now finding limit of (x-2)(x+3)/(x-2)
you can cancel out the x-2 from both the numerator and denominator, you just need to find the limit of x+3 as x approaches 2, which would just be 5
by
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L'Hopital says it's 5.
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Factor the numerator as (x+3)(x-2) then cancel out x-2 from the numerator and denominator you'll be left out with x+3 whose limit is 2+3=5 when x tends to 2.
So the answer is 5.
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Can you even have undefined limits? (genuine question)
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One way to intuitively understand the concept of limits is to plug in values very very close the number in question ,when faced with an indeterminate form ; eg plugging in 2.001 , 2.0001, 2.00001 and from the other side with 1.999, 1.9999, 1.99999 , when plugged in you’ll see it approaching 5 ; although in this particular problem factoring should be the first step as it cancels out the (x-2) in the denominator
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5, since it ends up in the indeterminate form if you plug in, you can use LHopital’s rule and take the derivative of the numerator and denominator, giving you (2x+1)/1. From there just plug in 2
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5 is the correct answer.
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5.
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its 5. There are two ways!  
1. Factoring the numerator. In this case its (x+3)(x-2), then u can cancel out x-2, leaving x+3 left. Just substitute the given limit (2), 2+3 = 5.  
2. By L'Hopital's Rule, since the equation is undefined, u can get the derivative of both the numerator and denominator. Equations becomes (2x + 1)/(1) or basically just 2x +1. Substitute the 2 again, and u'll get 5. (NOTE: There are certain rules when using the L'Hopital's. It can be only be used if the equation is 0/0 or infinity/infinity. If it's not in that form, there are ways to transform which i won't discuss here.)  


HOPE THIS HELPS!

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