The problem and my work towards solving it are attached. But not very sure about the answer. Can anybody please help.
Thanks.
The problem and my work towards solving it are attached. But not very sure about the answer. Can anybody please help.
Thanks.
Note that the as x → 1, the denominator → 0.
“... mathematics is only the art of saying the same thing in different words” - B. Russell
Yeah. That is the obvious problem.
Then how to find the possible values of a. Is it what I indicated i.e. a>1.
Please think about the limit existing at all. Factor the denominator.
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
moremathlearner already did this.
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"English is the most ambiguous language in the world." ~ Yours Truly, 1969
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.
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
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?
. . . . .[tex]\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}[/tex]
[tex]\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}\, >\, 0[/tex]
[tex]\mbox{Factorize den }\, \longrightarrow\, (2x\, +\, 1)(x\, -\, 1)[/tex]
[tex](x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0[/tex]
________________________________
. . . . . . .[tex][(4x^2\, -\, 1)\, (x^2\, -\, 1)]\, \nearrow[/tex]
[tex](x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0[/tex]
. . . . . . . . . . . . . . .[tex]\searrow\, \swarrow[/tex]
. . . . . . . . . . . . .[tex]\mbox{both}\, >\, 0[/tex]
[tex]\therefore (x^2\, -\, 2x\, +\, a)\, >\, 0[/tex]
[tex]\mbox{sub lim }\, x\, \rightarrow\, 1[/tex]
[tex]1\, -\, 2\, +\, a\, >\, 0[/tex]
[tex]-1\, +\, a\, >\, 0[/tex]
[tex]\underline{ a\, >\, 1 }[/tex]
Original poster: Please reply with corrections or confirmation. Thank you!
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