First step of this simple limit problem

[sachin_naik04]

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, please help me thanks in advance

 

[matticus]

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

 

[sachin_naik04]

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, please 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 please

 

[Mark44]

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).

 

[sachin_naik04]

@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

 

[sachin_naik04]

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

 

[HallsofIvy]

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.

 

[DarkOni]

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.

 

[Mentallic]

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

 

[sachin_naik04]

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 textbook 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?

 

[Mark44]

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.

 

[sachin_naik04]

ya thanks a lot