How Do You Find the Limit of A(z) as z Approaches 3?

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SUMMARY

The limit of the function A(z) = (2z - 6) / (z² - 5z + 6) as z approaches 3 can be determined by factoring both the numerator and denominator. The denominator factors to (z - 2)(z - 3), while the numerator can be factored to 2(z - 3). By canceling the (z - 3) term, the limit simplifies to 2, as z approaches 3.

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A(z)= 2z-6/ z2-5z+6

find the limit as z-->3

I factored the denominator to get (z-2)(z-3), and manipulated the function to get:

2z2-62/(z-2)(z-3)(2z+6)

but I'm just not sure what to do in order to get rid of the (z-3)

I hope there's some obvious answer that I'm not seeing!
 
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coolbeans33 said:
A(z)= 2z-6/ z2-5z+6

find the limit as z-->3

I factored the denominator to get (z-2)(z-3), and manipulated the function to get:

2z2-62/(z-2)(z-3)(2z+6)

but I'm just not sure what to do in order to get rid of the (z-3)

I hope there's some obvious answer that I'm not seeing!

You can factor the numerator to $$2(z-3)$$. Since z never reaches 3 you can cancel $$x-3$$
 

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