a=2 will indeed result in a finite limit in this case. If a = 2, then the numerator of your final limit will just be x-1. Then you can reduce the fraction to just 1/(x+1), and limit x --> 1 will exist by plugging in.
I'm having a bit of a difficult time understanding a problem that asks:
Find all values of such that
exists and is finite.
I feel that this should be simple, but I am stuck. The following is about as far I can make sense of it:
Thinking about numbers very close to 1, I believe a = 2; however, I am unable to work through it.
Any help in understanding what I am not seeing would be greatly appreciated.
Thank you in advance.
Take care.
/alan
a=2 will indeed result in a finite limit in this case. If a = 2, then the numerator of your final limit will just be x-1. Then you can reduce the fraction to just 1/(x+1), and limit x --> 1 will exist by plugging in.
If we use the direct substitution we get now for that to have an existing limit
We must have 2-a=0 the a=2
so we have so the limit exists
Now we can prove that this is the only solvation by LH rule
so in order for the limit to exist we must have the case so we can apply LH rule
So, is determining the value of really just evaluating
and then plugging in for to determine if the limit exists?
I apologize if this a dumb question. I want to be sure I fully understand every limit problem I do, and sometimes the solutions seem simpler to reach than expected.
Yea. If you think about it, at x = 1, the denominator will be 0 and there is nothing you can do about it. So you will have a limit of the form b/0, which is +-infinity, unless b is zero itself, because 0/0 is an indeterminate form. To keep the limit finite, you need to balance the infinitesimal size of the denominator by making the numerator infinitesimal as well, as x --> 1.