Find ##b-a## which satisfies following limit

[MatinSAR]

Homework Statement

Find $b-a$ which satisfies following limit:
$\lim_{x \rightarrow 1} {\frac {x-2} {x^3+ax+b}} = -\infty$

Relevant Equations

Please see below.

$\lim_{x \rightarrow 1} {\frac {x-2} {x^3+ax+b}} = -\infty$
The limit is equal to $\frac {-1} {1+a+b}$ .
so I can say that $a+b = -1$.
But I cannot find another equation to find both $b-a$.

 

[pasmith]

You need to make the denominator zero at x = 1, but you don't want it to change sign, or the limit from below will be -1 times the limit from above, and the two-sided limit will not exist. Thus (x - 1)^2 must be a factor of the denominator.

 

[MatinSAR]

You need to make the denominator zero at x = 1, but you don't want it to change sign, or the limit from below will be -1 times the limit from above, and the two-sided limit will not exist. Thus (x - 1)^2 must be a factor of the denominator.

Yes. I understand now. Thanks for your help.

 

[pasmith]

Setting <br /> \left. \frac{d}{dx}(x^3 + ax + b)\right|_{x=1} = 0 will give a directly, but fully factorising the denominator as above will confirm that the other linear factor is positive near x = 1.

 

[PeroK]

Homework Statement: Find $b-a$ which satisfies following limit:
$\lim_{x \rightarrow 1} {\frac {x-2} {x^3+ax+b}} = -\infty$
Relevant Equations: Please see below.

$\lim_{x \rightarrow 1} {\frac {x-2} {x^3+ax+b}} = -\infty$
The limit is equal to $\frac {-1} {1+a+b}$ .
so I can say that $a+b = -1$.
But I cannot find another equation to find both $b-a$.

This hypothesis of the question seems to fully specify $a$ and $b$?

 

[MatinSAR]

This hypothesis of the question seems to fully specify $a$ and $b$?

I'm not sure if I understand your question well.
The question asks for $b-a$ value. So we need to find both $a$ and $b$. I found $b-a = 5$.

 

[Mark44]

This hypothesis of the question seems to fully specify a and b?

The question asks for b−a value. So we need to find both a and b. I found b−a=5.

I don't believe it asks you to find both a and b, just the value of b - a.
If I understand @PeroK's comment/question correctly, he seems to be questioning whether both a and b need to be found.

In any case, how did you come up with b - a = 5?

 

[PeroK]

I don't believe it asks you to find both a and b, just the value of b - a.
If I understand @PeroK's comment/question correctly, he seems to be questioning whether both a and b need to be found.

Quite the reverse. Why ask for $b - a$ when we can find both $a$ and $b$? It suggested to me that any $a, b$ with a common difference would work.

 

[Mark44]

Quite the reverse. Why ask for b−a when we can find both a and b? It suggested to me that any a,b with a common difference would work.

That's probably the correct interpretation, although "find b - a which satisfies following limit" doesn't seem to be clearly stated. A better problem statement in this case would be "find a and b that satisfy the following limit."

 

[martinbn]

Quite the reverse. Why ask for $b - a$ when we can find both $a$ and $b$? It suggested to me that any $a, b$ with a common difference would work.

Yes, that is misleading, but i have seen many problems like that. They make you think that there is a clever way to find what is asked without the calculation for both numbers, just to be disappointed when you see the author's solution.

 

[MatinSAR]

@Mark44 @PeroK @martinbn What I have done to solve using @pasmith help :

Thus (x - 1)^2 must be a factor of the denominator.

 

[haruspex]

@Mark44 @PeroK @martinbn What I have done to solve using @pasmith help :

View attachment 336674

Yes, that works, but maybe this is a bit easier:
$(x-1)^2$ being a factor of $P(x)$ implies $x-1$ is a factor of $P'(x)$. So solve $3x^2+a=0$ for $x=1$.

 

[MatinSAR]

Yes, that works, but maybe this is a bit easier:
$(x-1)^2$ being a factor of $P(x)$ implies $x-1$ is a factor of $P'(x)$. So solve $3x^2+a=0$ for $x=1$.

I didn't know about this ...
It's far easier than what I have done!!! Thanks a lot for your help.