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

[Vali]

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!

 

[Dr.Peterson]

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.

Please show us the solution you got using the quadratic formula, so we can see whether your error is there, or in finding the limit. The limit is not particularly easy, so you may well be missing a detail there. We can't know how to help without seeing your work.

 

[Vali]

This is my try.

 

[Dr.Peterson]

This is my try.

Your solution for x₁ is correct. Go ahead and find the limit, showing your steps (it will take several). Note that the limit has the initial form ∞-∞, which is indeterminate.

My first step, after simplifying a little, was to "rationalize the numerator", multiplying top and bottom by the conjugate, which is one of the standard tricks for limits involving radicals.

 

[Vali]

Thank you again, I did it.I was lazy a little bit (I didn't try to simplify x1)
I replaced m with x because I usually work with x.I have one question.My teacher told us a way to say the limit without factor "x" (the last line).The numerator is 3x+1 so I have three "x".At the denominator I have one x + another x which comes from that sqrt(x^2..) so I have 2x and the limit is 3/2 (using that rule with coefficients and power of x).It's okay to approach the limit like this ?
Thank you very much, again!

 

[Dr.Peterson]

Thank you again, I did it.I was lazy a little bit (I didn't try to simplify x1)
I replaced m with x because I usually work with x.I have one question.My teacher told us a way to say the limit without factor "x" (the last line). The numerator is 3x+1 so I have three "x". At the denominator I have one x + another x which comes from that sqrt(x^2..) so I have 2x and the limit is 3/2 (using that rule with coefficients and power of x).It's okay to approach the limit like this ?
Thank you very much, again!

Excellent.

Your written work is just what I did.

The more informal approach is valid, as long as you are thinking carefully; if I were at all unsure, I would be sure to use the formal method. (Limits can be very dangerous places to use "magic".) A teacher may or may not accept it, depending on the level of formality of the course.