Limits and negative infinity

[tyro89]

Hello, my name is Erik and I’m looking for help regarding an explanation (or a proof) of why the following concept is true. (My teacher cannot give me one and neither can the calculus manual, they both simply state that this is just how it is)

problem:
img508.imageshack.us/img508/4498/problemap1.gif

I’m not sure I’m using the correct notation in solving this problem but really what I’m looking for is the proof or explanation why this is possible and why this is how it should be solved.

Thanks alot

Erik

[HallsofIvy]

Hello, my name is Erik and I’m looking for help regarding an explanation (or a proof) of why the following concept is true. (My teacher cannot give me one and neither can the calculus manual, they both simply state that this is just how it is)

problem:
img508.imageshack.us/img508/4498/problemap1.gif

I’m not sure I’m using the correct notation in solving this problem but really what I’m looking for is the proof or explanation why this is possible and why this is how it should be solved.

Thanks alot

Erik
Why what is possible? There's nothing there that isn't pretty basic algebra

The problem is to find the limit, as x -> -infinity, of sqrt(4x^2+ x- 3}/(x-1).

A fairly standard way of handling fractions, as x goes to infinity or negative infinity, is to divide by the largest power of x- so that x winds up in the denominator: as x goes to infinity, 1/x goes to 0 which is much easier to handle.

Here, they divide both numerator and denominator by -x. Do you want "proof" that dividing both numerator and denominator of a fraction by the same thing does not change the value of the fraction? Surely you learned that long ago! The only reason for the "-" is that, since x itself will be negative, -x= |x|.

sqrt{4x^2+ x- 3}/(-x)= sqrt{(4x^2+ x- 3)/x^2}. Taking the "-x" inside the square root it becomes "x^2" because sqrt{x^2}= |x|= -x.
sqrt{(4x^2+ x- 3)/x}= sqrt{4+ 1/x- 3/x^2} and that goes to 4 as x goes to -infinity because the fractions go to 0.

In the denominator, (x-1)/-x= -x/x+ 1/x= -1+ 1/x. Again, because 1/x goes to 0 as x goes to -infinity, that goes to -1.

Since both numerator and denominator HAVE limits, the limit of the fraction is 4/-1= -4.