Limits

[wassupman]

Hello, Can anyone point me in the right direction to start this problem? Do I factor the numerator?

 

[tkhunny]

This is why we study algebra and rational functions.

1) Can you factor the numerator?

2a) If so, does it have a factor of (x-5)?
2b) If not, what do we know about vertical asymptotes?

 

[Steven G]

These are the steps in my opinion for limits.

1) Plug in the value that x is approaching.
2) If you get a real number, then you are done. The answer is that real number.
3) if you get a non-zero number over zero, then you are done. The limit does not exist.
4) If you get 0/0, then you are not done. Somehow you have to get rid of that 0/0. Be clever. Then proceed back to step1.

I would never bother to factor before direct substitution. Did I say never?

 

[pka]

$\large x^3-9x^2+21x-5=(x-5)(x^2-4x+1)$

 

[JeffM]

These are the steps in my opinion for limits.

1) Plug in the value that x is approaching.
2) If you get a real number, then you are done. The answer is that real number.
3) if you get a non-zero number over zero, then you are done. The limit does not exist.
4) If you get 0/0, then you are not done. Somehow you have to get rid of that 0/0. Be clever. Then proceed back to step1.

I would never bother to factor before direct substitution. Did I say never?

Jomo

This practical advice will not necessarily work for piecewise functions, which people who write textbooks love to throw at students who are learning about continuity and limits.

 

[tkhunny]

1) Plug in the value that x is approaching.

Personally, I would never recommend this, for two reasons:
1) It's a limit. Substitution is not acceptable until after one has demonstrated continuity. (Not everyone agrees with me.)
2) Too many calculus students need the algebra practice. May as well get it now! There are prerequisites for a reason.

 

[LarlSangan42]

Assume it can be factored so (x-5)*(a*x^2+b*x+c)=x^3-9*x^2+21*x-5 then you can find a,b,c easily
ax^3=x^3 => a=1 ,
(-5*a+b)*x^2= -9*x^2 => b=-4 etc.

Also after you find your result(the limit)

you can find G(4.999) which should be almost the limG(x) (x→5). To see if you are correct.

you can also calculate G(5.111) instead

If G(5.111) is very different from G(4.999) the limit doesn't exist.

Well that's just for the confirmation that you are right , not the analytical solution. So don't show them to your math teacher he might freak out.. lol.

Yeah L'Hopital is the quick way if you have learned derivatives

 

[yoscar04]

Lazy people would use L'Hopital (in case you studied it already). Of course a more basic approach would be to factorize as some of the experienced forum members had suggested to you.

 

[HallsofIvy]

As Jomo said, first try putting the limit value for x, here 5, into the fraction. The only "difficult part" is the last- if both numerator and denominator are 0, where he said "4) If you get 0/0, then you are not done. Somehow you have to get rid of that 0/0. Be clever." Here is a hint that will help you be clever: If x= a makes a polynomial 0 then x- a is a factor of the polynomial!
Here the polynomial in the numerator is $\displaystyle x^3- 9x^2+ 21x- 5$ and the denominator is x- 5. Clearly when x= 5, the denominator is 5- 5= 0. According to what Jomo said, if the numerator is NOT 0 when x= 5, there is no limit. But setting x= 5 in the numerator gives 125- 225+ 105- 5= 100- 100= 0. So we have to be "clever". Fortunately we now know that x- 5 is a factor of $\displaystyle x^3- 9x^2+ 21x- 5$. x divides into $\displaystyle x^3$ $\displaystyle x^2$ times and $\displaystyle x^2(x- 5)= x^3- 5x^2$. Subtracting $\displaystyle x^2(x- 5)$ from $\displaystyle x^3- 9x^2+ 21x- 5$ leaves a remainder of $\displaystyle -4x^2+ 21x- 5$. x divides into that $\displaystyle -4x$ time. $\displaystyle -4x$ times x- 5 is $\displaystyle -4x^2+ 20x$. Subtracting $\displaystyle -4x^2+ 20x$ from $\displaystyle -4x^2+ 21x- 5$ leaves a remainder of $\displaystyle x- 5$ and $\displaystyle x- 5$ obviously divides into that once with no remainder!
$\displaystyle x^3- 9x^2+ 21x- 5= (x- 5)(x^2- 4x+ 1)$ so $\displaystyle \frac{x^3- 9x^2+ 21x- 5}{x- 5}= \frac{(x- 5)(x^2- 4x+ 1)}{x- 5}$.
As long as $\displaystyle x\ne 5$ we can cancel "x- 5" in the denominator leaving $\displaystyle x^2- 4x+ 1$. And it is important to remember that the limit, $\displaystyle \lim_{x\to a} f(x)$ depends only on the values of f(x) close to a but not at x= a! This limit is the same as $\displaystyle \lim_{x\to 5} x^2- 4x+ 1$. And now we can do #1 on Jomo's list. Setting x= 5, $\displaystyle 5^2- 4(5)+ 1= 25- 20+ 1= 1+ 1= 2$.
$\displaystyle \lim_{x\to 5}\frac{x^3- 9x^2+ 21x- 5}{x- 5}= \lim_{x\to 5} x^2- 4x+ 1= 2$.
(If the given function is defined "piecewise", as JeffM seems to be concerned about, Jomo's method can be applied to each "piece", again recalling that $\displaystyle \lim_{x\to a} f(x)$ depends only on what happens close to a, so if a is in one "piece" we can ignore all the others and if a is the point where one "piece" meets another, we can just look at those two pieces.)

 

[yoscar04]

I recall that at last year of High School we were taught a method called Ruffini technique for factorization of polynomials. I am not sure if it is still done. In any event, it may come in handy for working out this kind of limits.

 

[JeffM]

@HallsofIvy

Yes, that is what I was concerned about, and it is exactly the kind of thing thrown at students on tests about continuity and limits. What you forgot to say is that after you evaluate the two pieces at the join separately, you must check that they are equal. Mere existence will not do.

My criticism about what Jomo said is not that it will not work perfectly for most real world applications or that Jomo himself will not recognize the exceptions. My criticism is that a situation very likely to arise on tests is not mentioned.

 

[LarlSangan42]

I recall that at last year of High School we were taught a method called Ruffini technique for factorization of polynomials. I am not sure if it is still done. In any event, it may come in handy for working out this kind of limits.

We did something called Horner's method or Horner's table. Idk if it's the same

 

[yoscar04]

Interesting, by the way it reminds me that the Russian changed the names of most of the famous theorems, like Gauss, Stokes by Russian names (Ostrogradsky, etc.).
This is Ruffini's method: https://mathworld.wolfram.com/RuffinisRule.html

 

[HallsofIvy]

Of course, Russians claim it is we who change the names!

 

[Singleton]

The given rational function, call it f(x) is undefined at x=5. When you evaluate a limit, it doesn't matter if the function is continuous or not. As you can see, it need not even be defined there. Even you are given say f(5)=12, you'd ignore it in evaluating the limit. (It would be a trick question).

You plug in x=5 in the numerator as Jomo said, to see what is going on. And then yes, you factor it as yourself suggested. By now you have seen the solution. Here's another way to handle the cubic:

x³ – 9 x² + 21 x – 5
= x (x² – 9x + 20 + 1) – 5
= x [ (x–4)(x–5) + 1 ] – 5 = ... algebra ...
= (x–5) [x(x–4) + 1]

The cancellation of x-5 top/bottom is valid because x is not 5. So the limit is 5(5–4)+1 = 6.