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

In summary, the task is to determine the value of \( b - a \) that satisfies a given limit condition.
  • #1
MatinSAR
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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##.
 
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  • #2
You need to make the denominator zero at [itex]x = 1[/itex], 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 [itex](x - 1)^2[/itex] must be a factor of the denominator.
 
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  • #3
pasmith said:
You need to make the denominator zero at [itex]x = 1[/itex], 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 [itex](x - 1)^2[/itex] must be a factor of the denominator.
Yes. I understand now. Thanks for your help.
 
  • #4
Setting [tex]
\left. \frac{d}{dx}(x^3 + ax + b)\right|_{x=1} = 0[/tex] will give [itex]a[/itex] directly, but fully factorising the denominator as above will confirm that the other linear factor is positive near [itex]x = 1[/itex].
 
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  • #5
MatinSAR said:
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##?
 
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  • #6
PeroK said:
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 ##.
 
  • #7
PeroK said:
This hypothesis of the question seems to fully specify a and b?

MatinSAR said:
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?
 
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  • #8
Mark44 said:
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.
 
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  • #9
PeroK said:
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."
 
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  • #10
PeroK said:
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.
 
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  • #11
@Mark44 @PeroK @martinbn What I have done to solve using @pasmith help :
pasmith said:
Thus [itex](x - 1)^2[/itex] must be a factor of the denominator.
1701760887860.png
 
  • #12
MatinSAR said:
@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##.
 
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  • #13
haruspex said:
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.
 

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

What does the notation ##b-a## represent in the context of limits?

In the context of limits, ##b-a## typically represents the difference between two constants or variables, where ##b## and ##a## are specific values that need to be determined based on the given limit condition. The goal is to find this difference that satisfies the limit equation.

How do you approach solving a limit problem to find ##b-a##?

To solve a limit problem to find ##b-a##, you usually start by analyzing the given limit expression. This involves simplifying the expression if possible, applying limit laws, and sometimes using techniques like L'Hôpital's Rule, factoring, or substitution. The aim is to isolate the variables ##b## and ##a## and determine their values such that the limit holds true.

What are some common techniques used to evaluate limits?

Common techniques to evaluate limits include direct substitution, factoring, rationalizing, using conjugates, applying L'Hôpital's Rule, and exploiting trigonometric identities. These methods help in simplifying the limit expression to a form where the limit can be directly computed or inferred.

Can you provide an example of a limit problem requiring the determination of ##b-a##?

Sure, consider the limit problem: Find ##b-a## such that
\(\lim_{x \to 0} \frac{\sin(bx) - \sin(ax)}{x} = 1\).
To solve this, we can use the small-angle approximation for sine, \(\sin(kx) \approx kx\) when \(x \to 0\). This simplifies the expression to \(\lim_{x \to 0} \frac{bx - ax}{x} = 1\), which further simplifies to \(b-a = 1\).

Why is understanding limits important in calculus?

Understanding limits is crucial in calculus because they form the foundation for defining derivatives and integrals. Limits help in understanding the behavior of functions as they approach specific points or infinity. They are essential for analyzing continuity, evaluating instantaneous rates of change, and solving problems involving optimization and motion.

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