If limx→−∞ x2−x+1−ax−b=0, is - Infinity Learn by Sri Chaitanya

A
$a = 1, b = \frac{1}{2}$
B
$a = 1, b = - \frac{1}{2}$
C
$a = - 1, b = \frac{1}{2}$
D
none of these

Solution:

We have,

$\begin{array}{l} lim_{x \to \infty} \{\sqrt{x^{2} - x + 1} - (a x + b)\} = 0 \\ \Rightarrow lim_{x \to \infty} \frac{x^{2} - x + 1 - (a x + b)^{2}}{\sqrt{x^{2} - x + 1} + (a x + b)} = 0 \\ \Rightarrow lim_{x \to \infty} \frac{x^{2} (1 - a^{2}) - x (1 + 2 a b) + 1 - b^{2}}{\sqrt{x^{2} - x + 1} + a x + b} = 0 \\ \Rightarrow 1 - a^{2} = 0 and 1 + 2 a b = 0 \\ \Rightarrow a = \pm 1 and b = \mp 1 / 2 . \end{array}$

For $a = - 1$ and $b = 1 / 2$ we observe that

$lim_{x \to \infty} (\sqrt{x^{2} - x + 1} - a x - b) = lim_{x \to \infty} (\sqrt{x^{2} - x + 1} + x - \frac{1}{2}) \to \infty$

Hence, $a = 1$ and $b = - \frac{1}{2}$

If $lim_{x \to - \infty} \{\sqrt{x^{2} - x + 1} - a x - b\} = 0$ , is

Solution:

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