First step of this simple "limit" problem 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
Re: First step of this simple "limit" problem l'hospital's rule applies nicely. if you haven't learned that, then just factor the numerator and denominator.
Re: First step of this simple "limit" problem 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
Re: First step of this simple "limit" problem The denominator factors into (x^{2} - 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).
Re: First step of this simple "limit" problem @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
Re: First step of this simple "limit" problem 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
Re: First step of this simple "limit" problem The fact that Mark44 wrote that in terms of [itex]x^2- 4[/itex] rather than x- 2 implies that he first recognized that [itex]x^4- 8x^2+16= (x^2)^2- 2(4)x^2+ 4^2[/itex] is of the form "[itex]u^2- 2au+ a^2[/itex]", 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.
Re: First step of this simple "limit" problem 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.
Re: First step of this simple "limit" problem 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 ([itex]x^4[/itex]) but you can treat it as a quadratic in some other variable. [tex]x^4-8x^2+16[/tex] if you let [itex]x^2=u[/itex] you then have: [tex]u^2-8u+16[/tex] which is a quadratic. Factoring this gives you: [tex](u-4)^2[/tex] And then you can substitute [itex]u=x^2[/itex] 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
Re: First step of this simple "limit" problem well i understood the problem,thanks everyone a lot now i have one more problem for the following simple limit lim.........3x(x^{2}-7x+6) x->3......(x+2)(x^{3}+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?
Re: First step of this simple "limit" problem 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.