You are not logged in.
lim (2- √(6-x))/(x-2)
x->2
lim (3x²-x+1)/(x+3)
x->-3-
lim (√(x+Δx)- √(x))/Δx
Δx->0
lim (sin3x)/x
x->0
lim (x+2)/(x³+8)
x->-2
Without graphing
Last edited by fusilli_jerry89 (2006-10-22 18:38:31)
Offline
1)
(for all x not equal to 2)Hence:
Last edited by gnitsuk (2006-10-22 21:43:42)
Offline
3)
for all dx not equal to zero.Hence:
Last edited by gnitsuk (2006-10-22 21:43:54)
Offline
4)
For this we need to make use of the important result:
Let
thenLast edited by gnitsuk (2006-10-22 22:03:52)
Offline
5)
for all x not equal to -2.Hence:
Offline
2)
Here the limit is not finite. As x approaches -3 from -'ve infinity the given function approaches -infinity
so:
Last edited by gnitsuk (2006-10-22 23:00:15)
Offline
L'Hopital's rule would be very useful in some of those.
It states that if f(x) and g(x) are both undefined for some value of x (call it p), then:
.Why did the vector cross the road?
It wanted to be normal.
Offline
2)
Here the limit is not finite. As x approaches -3 from -'ve infinity the given function approaches -infinity
so:
when I graph it, it looks like as it approaches -3 from the negative side, it rises to positive infinite, not negative.
Offline
Here you can see that as x tends to -3 from -infinity the value of the function is heading to -infinity.
You might have entered the function incorrectly?
Mitch.
Last edited by gnitsuk (2006-10-25 21:54:52)
Offline
oh woops, my window setting were too small, i didnt see the other part of the function.
Offline