SOLUTION: Do you think you could help me find these limits? Really appreciate your help, in advance. Thank you so much! X + 1 lim = x→

            
lim (x+1) = 		           
x -> -1  

We think of substituting for x numbers which are
very NEAR to -1 into (x+1).

Since when we substitute the number -1 into (x+1),
we get (-1+1) or 0, not .

So we can reason that when we substitute numbers very
close to, but not quite equal to -1 into (x+1), we
will get numbers very close to, but not quite equal,
to 0.  So 0 is the limiting value, so we say

lim (x+1) = 		           
x -> -1

--------------------------------

lim 3 = 		           
x->0  

Since "3" contains no values of x, we add 0x
to 3 and get

lim (3+0x) = 		           
x->0

We think of substituting for x numbers which are
very NEAR to 0 into (3+0x).

Since when we substitute the number 0 into (3+0x),
we get (3+0) or 3, which is not .

So we can reason that when we substitute numbers very
close to, but not quite equal to 0 into (3+0x), we
will get numbers very close to, but not quite equal,
to 0.  So 0 is the limiting value, so we say

lim 3 = 3 		           
x -> 0

For other problems when the function is just a constant
we know that any limit of it are just that constant.

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I can't make that one out.  Sorry. Repost it.
            (   x 
lim    		  =  		   
x→0     (   2 
	
Edwin