Question

I forgot how to do this! Can some one teach me how to do the process again?

Answer 1

\[\lim_{x \rightarrow 3} \frac{ x-3 }{ x^2-4x+3 }\]

Answer 2

factor the denominator

\[\lim_{x \rightarrow 2}\frac{ x-3 }{ (x-3)(x+1) }\]

Answer 3

x-3 cancels

Answer 4

1/x+1 = 1/2+1 = 1/3

\[\lim_{x \rightarrow 2}=\frac{ 1 }{ 3 }\]

Answer 5

x is aprroachs to 3 not 2. @zpupster

Answer 6

then 1/4

Answer 7

I'm sorry your right x--> 3

Answer 8

Yes. Let x approach 3 in the expression \[\frac{ 1 }{ x+1 }.\]

The result is the desired limit.