<p>Hello everyone, there are a handful of problems here, so any help would be appreciated. I just forgot these simple rules from the beginning of the year, so can anyone do these for me to jog my memory? Again, feel free to answer or not answer, but I would really like some help.</p>
<p>1) What is f(x) given Lim (x approaches 3[from positive direction])
f(x)= x^2, x<3
2x-3, x>3
2) Let f(x) = (x^2) - 9<br>
_________ x doesnt equal 3
x - 3
and 5, x = 3</p>
<p>List which of the following are true
1) lim (x approaches 3) f(x) exists
2) f(3) is defined
3) f is continuous at x = 3</p>
<p>1) lim as x->3+ (f(x)), just plug in 3 for f(x). Since your coming in from the right, plug it into the function that approaches 3 from teh right. Which would be 2x-3. 2(3)-3 = -4.</p>
<p>2) Don't exactly know what is being asked. I'm assuming, if x does not equal 3, f(x) = (x^2)-9, and if x = 3, f(x) = 5?
In that case, plug in 3 in the first equation. While f(3) = 5, the value of f(x) as x approaches 3 is NOT 5. As such, when you plug in 3 into the first equation, you get 0, which is the answer. Because the function APPROACHES zero, even though the value of the function is not zero at x=3. (Do you sorta get what I'm saying here?)</p>
<p>List following, blah blah...</p>
<p>1) for problem (1), no, because the right-hand and left-hand limits are not equivalent. for problem (2), yes, because the left/right hand limits are the same are the same, even though the function is discontinuous at x=3.</p>
<p>2) for problem (1), no (it's either > or < 3 as defined). for problem (2) yes</p>
<p>3) for problem (1), no. for problem (2), no. Just plug in x=3 and see for yourself.</p>