y=2+(1/x-3) lim x ->0 and infinitive I got the answer = 1.6 and infinitive. it is continuous. is this right?
is a bit unclear. y = 2 + [1/(x - 3)]?
the function \[y =2 + [1/(x-3)]\] is a rational function, so it's continuous only when it's denominator doesn't equal 0. (x-3) equals 0 if x = 3. Since you're looking for the limit as x-> 0, and x is continuous at that point, you can directly plug in the 0 into the equation, getting: 2+(1/0-3)) = 2+ -1/3 = 5/3. So the limit as x-> 0 of the function is 5/3.
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