
#1
Dec2809, 12:14 PM

P: 13

Lim.............X^33x^2+4
x>2........... X^48x^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 



#2
Dec2809, 12:26 PM

P: 107

l'hospital's rule applies nicely. if you haven't learned that, then just factor the numerator and denominator.




#3
Dec2809, 10:30 PM

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 



#4
Dec2809, 10:40 PM

Mentor
P: 20,962

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).




#5
Dec2809, 10:47 PM

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 



#6
Dec2809, 10:50 PM

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




#7
Dec2909, 05:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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. 



#8
Dec2909, 06:43 AM

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.




#9
Dec2909, 08:48 AM

HW Helper
P: 3,436

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^48x^2+16[/tex] if you let [itex]x^2=u[/itex] you then have: [tex]u^28u+16[/tex] which is a quadratic. Factoring this gives you: [tex](u4)^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 (x2) 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 straightforward from there 



#10
Dec2909, 09:23 AM

P: 13

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? 



#11
Dec2909, 11:30 AM

Mentor
P: 20,962

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.




#12
Dec3009, 08:03 AM

P: 13

ya thanks a lot



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