a limit question below

Question

anonymous · Answer

$$\lim_{x \rightarrow - \infty}\frac{ x-3 }{ \sqrt{x^2-5}-2 }$$

anonymous · Answer

first step is to rationalize the polynomial.  multiply the numerator and denominator by the conjugate: $$\frac{ \sqrt{x ^{2}-5}+1 }{ \sqrt{x ^{2}-5}+1 }$$

anonymous · Answer

you should get $$\frac{ (x-3 )(\sqrt{x ^{2}-5}+1}{(x-3)(x+3)  }$$

anonymous · Answer

x-3 cancels in numerator and denominator

ganeshie8 · Answer

$\large \lim_{x \rightarrow - \infty}\frac{ x-3 }{ \sqrt{x^2-5}-2 } $

$\large \lim_{x \rightarrow - \infty}\frac{ x-3 }{ \sqrt{x^2-5}-2 } \times \frac{\sqrt{x^2-5}+2}{\sqrt{x^2-5}+2} $

$\large \lim_{x \rightarrow - \infty}\frac{ (x-3)(\sqrt{x^2-5} + 2) }{ x^2-5-4 }  $

$\large \lim_{x \rightarrow - \infty}\frac{ (x-3)(\sqrt{x^2-5} + 2) }{ (x-3)(x+3) }  $

$\large \lim_{x \rightarrow - \infty}\frac{ \sqrt{x^2-5} + 2}{ x+3}  $

ganeshie8 · Answer

say y = -1/x => x = -1/y

as x-> -infty, y->0

ganeshie8 · Answer

$\large \lim_{y \rightarrow 0}\frac{ \sqrt{\frac{1}{y^2}-5} + 2}{ \frac{-1}{y}+3}  $

$\large \lim_{y \rightarrow 0}\frac{ y \times (\sqrt{\frac{1}{y^2}-5} + 2)}{ y \times (\frac{-1}{y}+3)}  $

$\large \lim_{y \rightarrow 0}\frac{  \sqrt{1-5y^2} + 2y}{ -1+3y}  $

$\large \frac{  \sqrt{1-0} + 0}{ -1+0}  $

$\large  -1  $