If lim [x->infinity] [sqrt(x^2+x+1)-a*x-b] =0

[hahalalamummy]

I haven't learned L'Hospital's rule yet, just simple solution

 

[pappus]

I haven't learned L'Hospital's rule yet, just simple solution

1. I assume that you know how to determine an asymptote.

2. Rewrite the equation:

\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{x^2+x+1} - ax - b = 0 \right)}\)

to:

\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{\left(x+\frac12\right)^2+\frac34} - ax - b = 0 \right)}\)

3. This equation is only true if a = 1 and \(\displaystyle \displaystyle{b = \frac12}\)

 

[HallsofIvy]

Seeing that square root, I would be inclined to "rationalize". That is, multiply by \(\displaystyle \frac{\sqrt{x^2+ x+ 1}+ ax+ b}{\sqrt{x^2+ x+ 1}+ ax+ b}\) to get \(\displaystyle \frac{x^2+ x+ 1- (ax+ b)^2}{\sqrt{x^2+ x+ 1}+ ax+ b}= \frac{(1- a)x^2+ (1- 2ab)x+ 1- b^2}{\sqrt{x^2+ x+ 1}+ ax+ b}\).

In order that the limit, as x goes to infinity, be 0, the numerator must be of lower degree in x than the denominator. The denominator has degree 1 ([itex]\sqrt{x^2}= |x|[/itex]) so we must have 1- a= 0 and 1- 2ab= 0. From the first, a= 1 and then 1- 2ab= 1- 2b= 0 =gives b= 1/2.

 

[lookagain]

2. Rewrite the equation:

\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{x^2+x+1} - ax - b = 0 \right)}\)

to:

\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{\left(x+\frac12\right)^2+\frac34} - ax - b = 0 \right)}\)

The closing parentheses need to be placed here instead:


\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{x^2+x+1} - ax - b \right) = 0 }\)


to:


\(\displaystyle \displaystyle{\lim_{x \to \infty}\left(\sqrt{\left(x+\frac12\right)^2+\frac34} - ax - b \right) = 0}\)

 

[hahalalamummy]

Seeing that square root, I would be inclined to "rationalize". That is, multiply by \(\displaystyle \frac{\sqrt{x^2+ x+ 1}+ ax+ b}{\sqrt{x^2+ x+ 1}+ ax+ b}\) to get \(\displaystyle \frac{x^2+ x+ 1- (ax+ b)^2}{\sqrt{x^2+ x+ 1}+ ax+ b}= \frac{(1- a)x^2+ (1- 2ab)x+ 1- b^2}{\sqrt{x^2+ x+ 1}+ ax+ b}\).

In order that the limit, as x goes to infinity, be 0, the numerator must be of lower degree in x than the denominator. The denominator has degree 1 ([itex]\sqrt{x^2}= |x|[/itex]) so we must have 1- a= 0 and 1- 2ab= 0. From the first, a= 1 and then 1- 2ab= 1- 2b= 0 =gives b= 1/2.

If a<0, b<0, the denominator still = + infinity?

 

[lookagain]

If a<0, b<0, the denominator still = + infinity?

It is not possible for a < 0, b < 0, because that would make -ax - b positive, and combined
with the square root can never equal 0.

 

[hahalalamummy]

Seeing that square root, I would be inclined to "rationalize". That is, multiply by \(\displaystyle \frac{\sqrt{x^2+ x+ 1}+ ax+ b}{\sqrt{x^2+ x+ 1}+ ax+ b}\) to get \(\displaystyle \frac{x^2+ x+ 1- (ax+ b)^2}{\sqrt{x^2+ x+ 1}+ ax+ b}= \frac{(1- a)x^2+ (1- 2ab)x+ 1- b^2}{\sqrt{x^2+ x+ 1}+ ax+ b}\).

In order that the limit, as x goes to infinity, be 0, the numerator must be of lower degree in x than the denominator. The denominator has degree 1 ([itex]\sqrt{x^2}= |x|[/itex]) so we must have 1- a= 0 and 1- 2ab= 0. From the first, a= 1 and then 1- 2ab= 1- 2b= 0 =gives b= 1/2.

what about case of the numerator = 0
does not exist a and b?
and another cases?

 

[daon2]

what about case of the numerator = 0
does not exist a and b?
and another cases?

There are no a and b which make the numerator 0 for arbitrary large x (and that is what matters). Remember they are assumed invariant constants, and cannot depend on x.

You might find a similar, more challenging, exercise worthwhile:

\(\displaystyle \displaystyle \lim_{x\to\infty} \left[\sqrt{ax^2+bx+c} - (sx+t)\right]\)

For what constants a,b,c,s,t will this limit make sense? When will the limit exist? This encompasses your exercise with a=b=c=1.