If \( \lim_{x \to \infty} (\sqrt{x^2 - x + 1} - ax) = b \), then the ordered pair (a, b) is :
Show Hint
Limits of the form \(\lim_{x \to \infty} (\sqrt{ax^2+bx+c} - px)\) are common. A finite limit exists only if \(p = \sqrt{a}\). The technique to solve them is always to multiply by the conjugate to remove the indeterminate \(\infty - \infty\) form.
Step 1: Analyze the limit form.
As \(x \to \infty\), \(\sqrt{x^2 - x + 1} \approx \sqrt{x^2} = x\). The expression is of the form \(\infty - \infty\), which is an indeterminate form. For this limit to exist and be a finite number 'b', the highest power of x must cancel out. This requires \(ax\) to behave like \(x\), so we must have \(a=1\).
If \(a \neq 1\), the limit would be \(\infty\) or \(-\infty\). So, for a finite limit b to exist, \(a=1\). Step 2: Evaluate the limit with a=1.
With \(a=1\), the limit becomes:
\[ b = \lim_{x \to \infty} (\sqrt{x^2 - x + 1} - x) \]
To resolve the \(\infty - \infty\) form, we multiply by the conjugate:
\[ b = \lim_{x \to \infty} \frac{(\sqrt{x^2 - x + 1} - x)(\sqrt{x^2 - x + 1} + x)}{\sqrt{x^2 - x + 1} + x} \]
\[ b = \lim_{x \to \infty} \frac{(x^2 - x + 1) - x^2}{\sqrt{x^2 - x + 1} + x} \]
\[ b = \lim_{x \to \infty} \frac{-x + 1}{\sqrt{x^2 - x + 1} + x} \]
Now the form is \(\infty / \infty\). We can divide the numerator and the denominator by the highest power of x, which is x.
\[ b = \lim_{x \to \infty} \frac{\frac{-x}{x} + \frac{1}{x}}{\sqrt{\frac{x^2}{x^2} - \frac{x}{x^2} + \frac{1}{x^2}} + \frac{x}{x}} \]
\[ b = \lim_{x \to \infty} \frac{-1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} \]
As \(x \to \infty\), the terms \(1/x\) and \(1/x^2\) go to 0.
\[ b = \frac{-1 + 0}{\sqrt{1 - 0 + 0} + 1} = \frac{-1}{\sqrt{1} + 1} = \frac{-1}{2} \]
Step 3: Determine the ordered pair (a, b).
We found that \(a = 1\) and \(b = -1/2\).
So, the ordered pair (a, b) is \((1, -1/2)\).
