# First step of this simple limit problem

by sachin_naik04
Tags: limit, simple, step
 P: 13 Lim.............X^3-3x^2+4 x->2........... X^4-8x^2+16 In the above "limit" problem i would like to know only the very first important step, the later part of the problem i will manage to solve as its easy, only the first step is a bit tricky so what could be the first step of this problem actually i am a B.com. student, plz help me thanks in advance
 P: 107 l'hospital's rule applies nicely. if you haven't learned that, then just factor the numerator and denominator.
 P: 13 no i haven't learned any l'hospitals rule, but how do i factor the denominator as its x^4 i want to solve it without that l'hospital method, plz help oh i am hearing that l'hospitals rule is easy, i hope i will understand, so if possible u can also explain this rule to solve the first step of this problem or the other but atleast something plz
 Mentor P: 21,397 First step of this simple limit problem The denominator factors into (x2 - 4)2, which can be further factored into (x - 2)2(x + 2)2. The numerator can also be factored. Some obvious candidates are (x - 1), (x + 1), (x - 2), (x + 2), (x - 4), and (x + 4).
 P: 13 @Mark44 oh thanks a lot for your help really great i will try it on my book and see if i have any difficulty, if yes then i will come back here
 P: 13 but one more thing in order to get that (x^2 - 4)^2 have u used your own mind or have u done some rough work
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 The fact that Mark44 wrote that in terms of $x^2- 4$ rather than x- 2 implies that he first recognized that $x^4- 8x^2+16= (x^2)^2- 2(4)x^2+ 4^2$ is of the form "$u^2- 2au+ a^2$", the standard form of a perfect square. I would have done this in a completely different way: The only reason you had a problem with that fraction was because both numerator and denominator go to 0 as x goes to 2. That tells you immediately that each polynomial has x- 2 as a factor. Divide the polynomial by x- 2 to find the other factor. Repeat if necessary.
 P: 4 I think u should try learning L'Hospital's rule too.....all you have to do then is differentiate the numerator & denominator separately(only in cases: 0/0, ∞/∞, etc.) and then apply the limit.
 HW Helper P: 3,562 Actually I would've gone with a combination of both Mark44's and Hallsofivy's ideas. Firstly, looking at the denominator, yes it's a quartic ($x^4$) but you can treat it as a quadratic in some other variable. $$x^4-8x^2+16$$ if you let $x^2=u$ you then have: $$u^2-8u+16$$ which is a quadratic. Factoring this gives you: $$(u-4)^2$$ And then you can substitute $u=x^2$ back and factorize further as Mark44 has done. For the numerator it is a cubic, so the above method cannot be used but you can then factor out (x-2) since when you substitute x=2 into it, you end up with 0. This of course means x=2 is a root of the equation. After that you'll have a quadratic so it's straight-forward from there
 P: 13 well i understood the problem,thanks everyone a lot now i have one more problem for the following simple limit lim.........3x(x2-7x+6) x->3......(x+2)(x3+4x+3) now if u see the above problem and substitute the value of "x" that is "3" then u get the answer as -9/35 therefore the denominator doesn't go to "0" and hence the answer but the text book says the correct answer for this problem is 5/2, looks like we have to simplify the problem is it so, because i had heard that whenever the denominator does NOT go to "0" then we can directly substitute the value of x without simplifying the problem so the question is should i directly substitute the value of "x", as the denominator does not go to "0" or i have to simplify it, if yes then why?
 Mentor P: 21,397 Since the denominator doesn't approach 0 as x approaches 3, you can simply substitute 3 in both the numerator and denominator. The value I get is also -9/35. If your book gets a different value, make sure that you are working the same problem as in your book.
 P: 13 ya thanks a lot

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