# Thread: Finding unknown variable in a Limit problem

1. ## Finding unknown variable in a Limit problem

Thanks.

2. Originally Posted by moremathlearner
Thanks.

Note that the as x → 1, the denominator → 0.

3. Yeah. That is the obvious problem.

Then how to find the possible values of a. Is it what I indicated i.e. a>1.

4. Please think about the limit existing at all. Factor the denominator.

5. Originally Posted by tkhunny
Factor the denominator.

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6. Thanks. Browser wasn't working at my other installation.

moremathlearner... You did NOT do what I suggested. You only factored the denominator. How about the other part?

With a factor of (x-1) in the denominator, this limit certainly fails to exist unless it has that same factor in the numerator. Demonstrate that, first.

Note: When your problem is quadratic, why in the name of reason would you make it quartic? Don't do that.

7. Originally Posted by moremathlearner
For interested readers, the text in the images appears to be as follows:

4. For what value(s) of a is the following limit positive?

. . . . .$\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}$

$\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}\, >\, 0$

$\mbox{Factorize den }\, \longrightarrow\, (2x\, +\, 1)(x\, -\, 1)$

$(x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0$
________________________________

. . . . . . .$[(4x^2\, -\, 1)\, (x^2\, -\, 1)]\, \nearrow$

$(x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0$
. . . . . . . . . . . . . . .$\searrow\, \swarrow$
. . . . . . . . . . . . .$\mbox{both}\, >\, 0$

$\therefore (x^2\, -\, 2x\, +\, a)\, >\, 0$

$\mbox{sub lim }\, x\, \rightarrow\, 1$

$1\, -\, 2\, +\, a\, >\, 0$

$-1\, +\, a\, >\, 0$

$\underline{ a\, >\, 1 }$