Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to \infty }(\sqrt{x^2-x+1}-ax)=b$, then the ordered pair $(a,b)$ is:

Imagine you are standing on a vast, open plain, watching two runners sprinting toward the horizon. One runner is $\sqrt{x^2-x+1}$, and the other is $ax$.

As $x$ approaches infinity, both runners are moving incredibly fast, their speeds growing without bound. You are asked to find the gap between them, $b$, as they reach the horizon.

This is the essence of the limit:

This is a classic $\infty -\infty$ indeterminate form. We cannot simply subtract infinity from infinity; we need to find the precise balance where the growth rates match.

Rationalization

When we encounter square roots in limits, we use the surgical tool of rationalization. We treat the expression like a fraction with a hidden denominator of $1$.

By multiplying the numerator and denominator by the conjugate, $\sqrt{x^2-x+1}+ax$, we invoke the difference of squares identity: $(A-B)(A+B)=A^2-B^2$.

The numerator transforms from a messy radical into a simple quadratic:

Grouping the terms, we get:

Degree Analysis

Now, look at the denominator: $\sqrt{x^2-x+1}+ax$. As $x$ becomes massive, the $x^2$ inside the square root dominates, making the whole denominator behave like $x+ax$, which is proportional to $x$.

If our numerator still had an $x^2$ term, the entire fraction would grow linearly with $x$, shooting off to infinity. But we are told the limit is a finite value $b$.

This forces a constraint upon us: the coefficient of the $x^2$ term in the numerator must be zero. Thus:

We reject $a=-1$ because that would turn our subtraction into an addition, leading to $\infty +\infty$. So, $a=1$ is our only path forward.

With $a=1$ locked in, our expression simplifies to:

To find $b$, we divide both the numerator and the denominator by $x$:

As $x\to \infty$, the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ vanish into zero. We are left with:

The journey is complete. We have navigated the indeterminate form, rationalized the radical, balanced the degrees, and arrived at the elegant result of $(a,b)=(1,-\frac{1}{2})$.