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## #1 2006-10-22 18:38:06

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

### Limits

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)

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## #2 2006-10-22 21:11:57

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

1)

(for all x not equal to 2)

Hence:

Last edited by gnitsuk (2006-10-22 21:43:42)

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## #3 2006-10-22 21:39:04

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

3)

for all dx not equal to zero.

Hence:

Last edited by gnitsuk (2006-10-22 21:43:54)

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## #4 2006-10-22 21:49:15

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

4)

For this we need to make use of the important result:

Let

then

Last edited by gnitsuk (2006-10-22 22:03:52)

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## #5 2006-10-22 21:55:36

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

5)

for all x not equal to -2.

Hence:

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## #6 2006-10-22 21:59:55

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

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)

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## #7 2006-10-23 00:13:22

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Limits

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.

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## #8 2006-10-25 18:02:07

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

### Re: Limits

gnitsuk wrote:

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.

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## #9 2006-10-25 20:45:54

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

### Re: Limits

Plot of (3X^2-X+1)/(X+3)

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)

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## #10 2006-10-26 12:15:16

fusilli_jerry89
Member
Registered: 2006-06-23
Posts: 86

### Re: Limits

oh woops, my window setting were too small, i didnt see the other part of the function.

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