Derivative by definition. Complex function?

[stalker7d7]

Limit as x approaches infinity. Complex function?

I apologize, I mixed up the methods. I meant to say that we needed to find the limit as x approaches infinity.

So I took a test, did fine, but on it were a few problems that looked like this:

Lim_x->∞f(x) = (7+12x-5x²)/(3+2x-3x²) (not sure on the exact problems, but they looked like that)

When I look at this, I'm (still) at a complete loss of what to do. I can't remember ever seeing anything like this before.

 

[srmichael]

How do I deal with the negatives in the polynomials?

What do you mean by this???

 

[soroban]

Hello, stalker7d7!

Use the definition of the derivative: .\(\displaystyle \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

differentiate: .\(\displaystyle f(x) \:=\: \dfrac{7+12x-5x^2}{3+2x-3x^2}\)

Tell Professor deSade that no one ... NO ONE ... asks for
. . the derivative of a quotient by definition on an exam.

Even \(\displaystyle f(x) \:=\:\dfrac{2x+3}{x}\) would take an inordinate amount of time.

And he/she used two quadratics . . . with negative \(\displaystyle x^2\)-terms,
. . creating even more opportunities for errors.


Please let us know how many students got the right answer.

 

[stalker7d7]

Hello, stalker7d7!
Tell Professor deSade that no one ... NO ONE ... asks for
. . the derivative of a quotient by definition on an exam.

You're correct. I mixed it up in my mind. It was actually asking for the limit as x appraches infinity. Which still boggled my mind.


p.s. I'm a bit stressed out, can you tell? :-?

 

[pka]

You're correct. I mixed it up in my mind. It was actually asking for the limit as x appraches infinity. Which still boggled my mind.

Suppose we have two polynomials.
\(\displaystyle p(x) = \sum\limits_{k = 0}^m {p_k x^{m - k} } \;\& \,q(x) = \sum\limits_{k = 0}^n {q_k x^{n - k} } \) where \(\displaystyle p_0\cdot q_0\ne 0\).

Then \(\displaystyle \lim _{x \to \infty } \dfrac{{p(x)}}{{q(x)}} = \left\{ {\begin{array}{r} {p_0 /q_0 } \\
0 \\
{ \pm \infty } \\
\end{array}\begin{array}{l}
{,if\;m = n} \\
{,if\;m < n} \\
{,if\;m > n} \\
\end{array}} \right.
\)

So it is easy to see that your limit is \(\displaystyle \frac{-5}{-3}\).

 

[afrazer721]

I apologize, I mixed up the methods. I meant to say that we needed to find the limit as x approaches infinity.

So I took a test, did fine, but on it were a few problems that looked like this:

Lim_x->∞f(x) = (7+12x-5x²)/(3+2x-3x²) (not sure on the exact problems, but they looked like that)

When I look at this, I'm (still) at a complete loss of what to do. I can't remember ever seeing anything like this before.

The most powerful part of the numerator is -5x^2, and in the denominator is -3x^2. Thus:
Lim_x->∞f(x) = (7+12x-5x²)/(3+2x-3x²)=Lim_{x->∞-5x^2/-3x^2=}Lim<sub>x->∞-5/-3=5/3.
The limit is formed by the coefficients of the most powerful parts: -5 in the numerator and -3 in the denominator.
Hope this does some good.

"The minute u settle for less than u deserve, u get even less than u settled 4." Maureen Dowd, writer</sub>

 

[medicalphysicsguy]

m, n clarification

Suppose we have two polynomials.
\(\displaystyle p(x) = \sum\limits_{k = 0}^m {p_k x^{m - k} } \;\& \,q(x) = \sum\limits_{k = 0}^n {q_k x^{m - k} }\)

May I ask if you meant

\(\displaystyle p(x) = \sum\limits_{k = 0}^m {p_k x^{m - k} } \;\& \,q(x) = \sum\limits_{k = 0}^n {q_k x^{n - k} }\)

Thanks, mpg

 

[pka]

May I ask if you meant
\(\displaystyle p(x) = \sum\limits_{k = 0}^m {p_k x^{m - k} } \;\& \,q(x) = \sum\limits_{k = 0}^n {q_k x^{n - k} }\)

YES. Good catch.