Limit (x1) as m->infinity ( x1 - the lowest root )

In summary, you are trying to find the limit as m grows without bound, and you believe that the lower root is $x_1 = m+1 - \sqrt{m^2-m}$. You can take the limit as $m\to\infty$, and you will find that x1 approaches $-1/2$ as m grows without bound.
  • #1
Vali
48
0
I have the following equation: x^2 - 2(m+1)x + 3m + 1=0
Also, I know that x1 is the lowest root of this equation.
I need to solve lim (x1) as m->infinity
A. 1
B. 3/2
C. 0
D. -1/2
E. -1
I tried to solve the equation with the discriminant then to calculate the limit but didn't work.
Also, I think that because x1 is the lowest root and the function graphic is a parabola, I tink that -b/2a (the peak of parabola) > x1 but I don't see how this condition would help me.
Some ideas?
Thanks!
 
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  • #2
Vali said:
I have the following equation: x^2 - 2(m+1)x + 3m + 1=0
Also, I know that x1 is the lowest root of this equation.
I need to solve lim (x1) as m->infinity
A. 1
B. 3/2
C. 0
D. -1/2
E. -1
I tried to solve the equation with the discriminant then to calculate the limit but didn't work.
Also, I think that because x1 is the lowest root and the function graphic is a parabola, I tink that -b/2a (the peak of parabola) > x1 but I don't see how this condition would help me.
Some ideas?
Thanks!
You did right to start by solving the equation, and you probably found that the lower root is $x_1 = m+1 - \sqrt{m^2-m}$. The trick now is to make that into a fraction, multiplying and dividing by $m+1 + \sqrt{m^2-m}$ to get $$x_1 = \frac{\bigl( m+1 - \sqrt{m^2-m}\bigr)\bigl( m+1 + \sqrt{m^2-m}\bigr)}{m+1 + \sqrt{m^2-m}}.$$ Can you take it from there, to get the limit as $m\to\infty$?
 
  • #3
Yes, I replaced m with x because I usually work with x.
Thank you very much for your help :)
 

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  • #4
Vali said:
I need to solve lim (x1) as m->infinity

Hi Vali; do you mean "What number does x1 approach as m grows without bound"?
 
  • #5
greg1313 said:
Hi Vali; do you mean "What number does x1 approach as m grows without bound"?

Yes.Sorry if I didn;t use the correct words.
 
  • #6
Hey, no problem. The notation is unusual so I have asked for a clarification to benefit those who may not understand. Happy foruming! :)
 

FAQ: Limit (x1) as m->infinity ( x1 - the lowest root )

What does "limit (x1) as m->infinity ( x1 - the lowest root )" mean?

This notation represents the limit of a function as the independent variable approaches infinity. Specifically, it refers to the limit of the lowest root of the function as the independent variable approaches infinity.

How is the limit of the lowest root calculated?

The limit of the lowest root can be calculated by finding the limit of the function as the independent variable approaches infinity, and then finding the lowest root of that limit function.

Why is the limit of the lowest root important?

The limit of the lowest root can provide valuable information about the behavior of a function as the independent variable approaches infinity. It can also help to determine the end behavior of a function and identify any asymptotes.

Can the limit of the lowest root be negative?

Yes, the limit of the lowest root can be negative. This would indicate that the function approaches a negative value as the independent variable approaches infinity.

How is the limit of the lowest root used in real-world applications?

The limit of the lowest root can be used in various fields such as physics, engineering, and economics to model and predict the behavior of systems as they approach infinity. It can also be used to analyze the stability and convergence of numerical methods.

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