Find-all-asymptotes-of-the-function-y-x-x-2-2-

Question Number 126008 by bramlexs22 last updated on 16/Dec/20

F i n d a l l a s y m p t o t e s o f t h e

f u n c t i o n y = \frac{x}{\sqrt{x^{2} + 2}} .

Answered by liberty last updated on 16/Dec/20

H o r i z o n t a l a s y m p t o t e : y = \lim_{x \to \infty} \frac{x}{\sqrt{x^{2} + 2}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + 2 x^{- 2}}} = 1

a n d y = \lim_{x \to - \infty} \frac{x}{\sqrt{x^{2} + 2}} = \lim_{x \to - \infty} \frac{- 1}{\sqrt{1 + 2 x^{- 2}}} = - 1

s o y = - 1 a n d y = 1 a r e t h e h o r i z o n t a l a s y m p t o t e s

Answered by ebi last updated on 16/Dec/20

l e t f (x) = \frac{x}{\sqrt{x^{2} + 2}}

v e r t i c a l a s y m p t o t e s :

s i n c e t h e \sqrt{x^{2} + 2} > 0,

∴ n o v e r t i c a l a s y m p t o t e s a v a i l a b l e .

h o r i z o n t a l a s y m p t o t e s :

\underset{x \to + \infty}{l i m} (\frac{x}{\sqrt{x^{2} + 2}}) = \underset{x \to \infty}{l i m} (\frac{1}{\sqrt{1 + \frac{2}{x^{2}}}})

= \frac{1}{\sqrt{1 + 0}} = 1

\underset{x \to - \infty}{l i m} (\frac{x}{\sqrt{x^{2} + 2}}) = \underset{x \to \infty}{l i m} (- \frac{x}{\sqrt{x^{2} + 2}})

= \underset{x \to \infty}{l i m} (- \frac{1}{\sqrt{1 + \frac{2}{x^{2}}}}) = - \frac{1}{\sqrt{1 + 0}} = - 1

∴ h o r i z o n t a l a s y m p t o t e s, y = - 1 a n d y = 1 .

s l a n k a s y m p t o t e s :

y = m x + c

m = \frac{f (x)}{x} = \frac{1}{\sqrt{x^{2} + 2}}

\underset{x \to + \infty}{l i m} (\frac{1}{\sqrt{x^{2} + 2}}) = 0

\underset{x \to - \infty}{l i m} (\frac{1}{\sqrt{x^{2} + 2}}) = \underset{x \to \infty}{l i m} (- \frac{1}{\sqrt{x^{2} + 2}}) = 0

∴ n o s l a n k a s y m p t o t e s a v a i l a b l e .

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