First step of this simple limit problem

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In summary, the first step of the "limit" problem is to apply L'Hospital's rule or factor the numerator and denominator. In the problem provided, 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 using (x - 1), (x + 1), (x - 2), (x + 2), (x - 4), and (x + 4). In another problem, the denominator does not approach 0 as x approaches 3, so the value can be directly substituted without simplifying the problem.
  • #1
sachin_naik04
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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
 
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  • #2


l'hospital's rule applies nicely. if you haven't learned that, then just factor the numerator and denominator.
 
  • #3


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
 
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  • #4


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).
 
  • #5


@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


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


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


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


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 :smile:
 
  • #10


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?
 
  • #11


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


ya thanks a lot
 

1. What is the first step of a simple limit problem?

The first step of a simple limit problem is to determine the value of the limit by plugging in the given input value into the function and simplifying the expression.

2. How do I know if a limit problem is simple or not?

A limit problem is considered simple if the function in the limit expression can be evaluated by direct substitution without any other algebraic manipulations. If the function is not continuous at the given input value or involves indeterminate forms like 0/0 or ∞/∞, it is not a simple limit problem.

3. What is the difference between a simple limit problem and a complex limit problem?

A simple limit problem can be evaluated by direct substitution, whereas a complex limit problem requires additional algebraic manipulations, such as factoring, rationalizing, or using L'Hôpital's rule, to evaluate the limit.

4. Do all limit problems have a defined value?

No, not all limit problems have a defined value. Some limit expressions may approach ∞ or -∞, while others may have no limit at all.

5. Can a simple limit problem have a different value than its actual limit?

Yes, a simple limit problem can have a different value than its actual limit if the function is not continuous at the given input value. In this case, the limit does not exist, but the evaluated value may still be a valid output of the function.

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