Question

y=2+(1/x-3)
lim x ->0 and infinitive
I got the answer = 1.6 and infinitive. it is continuous. is this right?

Answer 1

is a bit unclear. y = 2 + [1/(x - 3)]?

Answer 2

yea

Answer 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.