The-following-image-shows-the-functiond-f-x-xe-1-1-x-and-g-x-x-1-Can-you-explain-as-to-why-as-f-x-that-f-x-g-x-

Question Number 4281 by Filup last updated on 07/Jan/16

The following image shows the functiond

f (x) = {x e}^{\frac{1}{1 - x}} and g (x) = x - 1

Can you explain as to why as ∣ f (x) ∣\to \infty,

that f (x) \to g (x) .

Commented by Filup last updated on 07/Jan/16

Commented by Filup last updated on 07/Jan/16

x \to \pm \infty, ∣ x ∣\to \infty

\lim_{x \to \infty} {x e}^{\frac{1}{1 - x}} = \infty e^{0} = \infty (1)

\lim_{x \to \infty} x - 1 = \infty (2)

(1) and (2) tend to infinity . Thus limits

are equal .

Commented by Yozzii last updated on 07/Jan/16

f (x) = {x e}^{\frac{1}{1 - x}} (x \neq 1)

f^{^{'}} (x) = e^{{(1 - x)}^{- 1}} + x \times {(1 - x)}^{- 2} e^{{(1 - x)}^{- 1}}

f^{^{'}} (x) = e^{\frac{1}{1 - x}} (1 + \frac{x}{{(x - 1)}^{2}})

f^{^{'}} (x) = e^{{(1 - x)}^{- 1}} (\frac{x^{2} - x + 1}{x^{2} - 2 x + 1})

f^{^{'}} (x) = e^{{(1 - x)}^{- 1}} (\frac{1 - \frac{1}{x} + \frac{1}{x^{2}}}{1 - \frac{2}{x} + \frac{1}{x^{2}}})

∴ l = \lim_{x \to \infty} f^{^{'}} (x) = \lim_{x \to \infty} {e^{\frac{1}{1 - x}} (\frac{1 - \frac{1}{x} + \frac{1}{x^{2}}}{1 - \frac{2}{x} + \frac{1}{x^{2}}})}

l = e^{\frac{1}{- \infty}} \frac{1 - 0 + 0}{1 - 0 + 0} = 1 = g^{^{'}} (x) f o r x > 1 .

S o, t h e g r a p h o f f (x) i s a p p r o x i m a t e l y

p a r a l l e l t o t h a t o f g (x) f o r s u f f i c i e n t l y

l a r g e v a l u e s o f x .

l n f (x) = l n x + \frac{1}{1 - x}

∴ \lim_{x \to 1^{+}} l n f (x) = l n 1 - \infty = - \infty

∴ \lim_{x \to 1^{+}} f (x) = e^{- \infty} = 0

S o, x = 1 i s t h e a p p r o x i m a t e r o o t o f f (x) = 0 .

A l s o, \lim_{x \to \infty} l n f (x) = \infty \Rightarrow f (x) \to \infty

w h i l e f^{'} (x) \to 1 . S o, a s t r a i g h t l i n e

a p p r o x i m a t e s f (x) f o r l a r g e x .

G i v e n t h a t f^{^{'}} (x) \to 1 a s x \to \infty f o r x > 1

a n d x = 1 i s a r o o t o f f (x) = 0, a n

a s y m p t o t e t o f (x) a s x \to \infty i s f o u n d t o

b e y - 0 = (1) (x - 1) \Rightarrow y = x - 1 w h i c h

i s t h e f u n c t i o n g (x) .

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