# What is the result of lim x → ∞ (-x-1)/((x-1)(x-Sqrt(x^2+x+1)) ?

Jul 21, 2018

$2$

#### Explanation:

lim x → ∞ (-x-1)/((x-1)(x-sqrt(x^2+x+1))

= lim x → ∞ (-x-1)/((x-1)(x-sqrt((x+ 1/2)^2+3/4))

$y = x + \frac{1}{2} q \quad x = y - \frac{1}{2}$

= lim y → ∞ (-y-1/2)/((y-3/2)(y - 1/2-sqrt(y^2+3/4))

= lim y → ∞ (-1-1/(2y))/((y-3/2)(1 - 1/(2y)-sqrt(1+3/(4y^2)))

By binomial expansion:

$\sqrt{1 + \frac{3}{4 {y}^{2}}} = 1 + \left(\frac{1}{2}\right) \frac{3}{4 {y}^{2}} + \mathbb{O} \left(\frac{1}{y} ^ 6\right)$

= lim y → ∞ (-1-1/(2y))/((y-3/2)(1 - 1/(2y)-1 - (3/(8y^2)) ))

= lim y → ∞ (1+1/(2y))/((y-3/2)( 1/(2y) + 3/(8y^2) ))

= lim y → ∞ (1+cancel(1/(2y)))/(1/2 + cancel(3/(8y)) - cancel(3/(4y)) - cancel(9/(16y^2)))

$= 2$

$2$

#### Explanation:

$\setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{- x - 1}{\left(x - 1\right) \left(x - \setminus \sqrt{{x}^{2} + x + 1}\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{- \left(x + 1\right) \left(x + \setminus \sqrt{{x}^{2} + x + 1}\right)}{\left(x - 1\right) \left(x - \setminus \sqrt{{x}^{2} + x + 1}\right) \left(x + \setminus \sqrt{{x}^{2} + x + 1}\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{- \left(x + 1\right) \left(x + \setminus \sqrt{{x}^{2} + x + 1}\right)}{\left(x - 1\right) \left({x}^{2} - {\left(\setminus \sqrt{{x}^{2} + x + 1}\right)}^{2}\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{- \left(x + 1\right) \left(x + \setminus \sqrt{{x}^{2} + x + 1}\right)}{\left(x - 1\right) \left({x}^{2} - {x}^{2} - x - 1\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{- \left(x + 1\right) \left(x + \setminus \sqrt{{x}^{2} + x + 1}\right)}{- \left(x - 1\right) \left(x + 1\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{x + \setminus \sqrt{{x}^{2} + x + 1}}{x - 1}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{x + x \setminus \sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2}}{x \left(1 - \frac{1}{x}\right)}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{1 + \setminus \sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2}}{1 - \frac{1}{x}}$
$= \setminus \frac{1 + \setminus \sqrt{1 + 0 + 0}}{1 - 0}$

$= \setminus \frac{2}{1}$

$= 2$