"Evaluate the folowing limits. limx→∞(x2−7x+1−x) limx→−∞(x2−7x+1−x)" solved and explained

Tobias Ali

Answered question

2021-10-16

Evaluate the folowing limits.

lim_{x \to \infty} (\sqrt{x^{2} - 7 x + 1} - x)

lim_{x \to - \infty} (\sqrt{x^{2} - 7 x + 1} - x)

Answer & Explanation

izboknil3

Skilled2021-10-17Added 99 answers

a) $lim_{x \to \infty} (\sqrt{x^{2} - 7 x + 1} - x) = lim_{x \to \infty} (\sqrt{x^{2} - 7 x + 1} - x) \times \frac{\sqrt{x^{2} - 7 x + 1} + x}{\sqrt{x^{2} - 7 x + 1} + x}$
$lim_{x \to \infty} \frac{(x^{2} - 7 x + 1) - x^{2}}{\sqrt{x^{2} - 7 x + 1} + x} = lim_{x \to \infty} \frac{- 7 x + 1}{\sqrt{x^{2} - 7 x + 1} + x}$
$= lim_{x \to \infty} \frac{- 7 + \frac{1}{x}}{\sqrt{1 - \frac{7}{x} + \frac{1}{x^{2}}} + 1}$ =-7
b) $lim_{x \to - \infty} (\sqrt{x^{2} - 7 x + 1} - x)$
let us take
x=-y
$x \to - \infty \Rightarrow y \to \infty$
Therefore $lim_{x \to - \infty} (\sqrt{x^{2} - 7 x + 1} - x) = lim_{y \to \infty} (\sqrt{{(- y)}^{2} - 7 (- y) + 1} - (- y))$
$lim_{x \to - \infty} (\sqrt{x^{2} - 7 x + 1} - x) = lim_{y \to \infty} (\sqrt{y^{2} + 7 y + 1} + y) = \infty$
Answer:
$lim_{x \to - \infty} (\sqrt{x^{2} - 7 x + 1} - x) = - 7$
$lim_{x \to \infty} (\sqrt{x^{2} - 7 x + 1} - x) = \infty$

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