You guessed it... a limit question.

I'm having a bit of a difficult time understanding a problem that asks:

Find all values of $\displaystyle {a}$ such that

$\displaystyle \lim_{x\to1}\left(\frac{1}{x-1}-\frac{a}{x^2-1}\right)$

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:

$\displaystyle \rightarrow \lim_{x\to1}\left(\frac{1}{x-1}-\frac{a}{(x-1)(x+1)}\right)$

$\displaystyle \rightarrow \lim_{x\to1}\left(\frac{x+1}{(x-1)(x+1)}-\frac{a}{(x-1)(x+1)}\right)$

$\displaystyle \rightarrow \lim_{x\to1}\left(\frac{x+1-a}{(x-1)(x+1)}\right)$

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

Re: You guessed it... a limit question.

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.

Re: You guessed it... a limit question.

If we use the direct substitution we get $\displaystyle \frac{2-a}{0}$ now for that to have an existing limit

We must have 2-a=0 the a=2

so we have $\displaystyle \lim_{x\to 1}\frac{x-1}{(x-1)(x+1)}=\frac{1}{2}$ 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 $\displaystyle \frac{0}{0}$ so we can apply LH rule

$\displaystyle \lim_{x\to 1}\frac{1}{2x}=\frac{1}{2}$

Re: You guessed it... a limit question.

So, is determining the value of $\displaystyle a$ really just evaluating

$\displaystyle x+1-a=0$

$\displaystyle \rightarrow (1)+1-a=0$

$\displaystyle \rightarrow 2-a=0$

$\displaystyle \rightarrow a=2$

and then plugging in $\displaystyle 2$ for $\displaystyle a$ 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.

Re: You guessed it... a limit question.

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.