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Question Number 73555 by Rio Michael last updated on 13/Nov/19

f i n d

a) \underset{x \to - \infty}{L i m} \frac{l n (1 + e^{x})}{e^{x}}

b) \underset{x \to + \infty}{l i m} (\sqrt{x^{2} + 3 x} - x)

c) \underset{x \to 0}{l i m} \sqrt{x} l n (s i n x)

d) \underset{x \to + \infty}{l i m} \frac{e^{2 x + 1} - e^{x}}{x^{2} - x - 1}

e) \underset{x \to - \infty}{l i m} [\sqrt{1 - {x e}^{x}}]

Commented by mathmax by abdo last updated on 13/Nov/19

a) w e u s e c h a n g . e^{x} = t \Rightarrow {l i m}_{x \to - \infty} \frac{l n (1 + e^{x})}{e^{x}} = {l i m}_{t \to 0} \frac{l n (1 + t)}{t}

= 1

b) w e h a v e \sqrt{x^{2} + 3 x} - x = \sqrt{x^{2} (1 + \frac{3}{x})} - x

= x \sqrt{1 + \frac{3}{x}} - x \sim x (1 + \frac{3}{2 x}) - x (x \to + \infty)

\sim \frac{3}{2} \Rightarrow {l i m}_{x \to + \infty} \sqrt{x^{2} + 3 x} - x = \frac{3}{2}

c) {l i m}_{x \to 0} \sqrt{x} l n (s i n x) ? w e h a v e s i n x \sim x \Rightarrow \sqrt{x} l n (s i n x) \sim \sqrt{x} l n (x)

a n d {l i m}_{x \to 0} \sqrt{x} l n (x) =_{\sqrt{x} = u} {l i m}_{u \to 0} u l n (u^{2}) = {l i m}_{u \to 0} 2 u l n (u) = 0

Commented by mathmax by abdo last updated on 13/Nov/19

d) l e t f (x) = \frac{e^{2 x + 1} - e^{x}}{x^{2} - x - 1} \Rightarrow f (x) = \frac{e^{2 x + 1}}{x^{2}} \times \frac{1 - e^{x - 2 x - 1}}{1 - \frac{1}{x} - \frac{1}{x^{2}}}

w e h a v e {l i m}_{x \to + \infty} \frac{e^{2 x + 1}}{x^{2}} = + \infty (e^{x} d e f e a t x^{n} \dots)

{l i m}_{x \to + \infty} \frac{1 - e^{- x - 1}}{1 - \frac{1}{x} - \frac{1}{x^{2}}} = 1 \Rightarrow {l i m}_{x \to + \infty} f (x) = + \infty

Commented by Rio Michael last updated on 14/Nov/19

t h a n k s s i r

Commented by malwaan last updated on 14/Nov/19

b) \underset{x \to \infty}{l i m} \frac{(\sqrt{x^{2} + 3 x} - x) (\sqrt{x^{2} + 3 x} + x)}{\sqrt{x^{2} + 3 x} + x}

= \underset{x \to \infty}{l i m} \frac{x^{2} + 3 x - x^{2}}{x [\sqrt{1 + \frac{3}{x}} + 1]}

= \underset{x \to \infty}{l i m} \frac{3}{\sqrt{1 + \frac{3}{x}} + 1} = \frac{3}{2}

Answered by ajfour last updated on 13/Nov/19

a) e^{- \infty} \to 0 a n d \lim_{h \to 0} \frac{\ln (1 + h)}{h} = 1

A n s . t o (a) = 1

b) \underset{x \to + \infty}{l i m} (\sqrt{x^{2} + 3 x} - x)

= \lim_{x \to \infty} \frac{3 x}{\sqrt{x^{2} + 3 x} + x} = \lim_{x \to \infty} \frac{3}{\sqrt{1 + \frac{3}{x}} + 1}

= \frac{3}{2}

c) \underset{x \to 0}{l i m} \sqrt{x} l n (s i n x)

= \frac{\lim_{x \to 0} \frac{d}{d x} \ln (\sin x)}{\lim_{x \to 0} \frac{d}{d x} (\frac{1}{\sqrt{x}})} = \frac{\lim_{x \to 0} \frac{1}{\tan x}}{\lim_{x \to 0} (- \frac{1}{2 x \sqrt{x}})}

= \lim_{x \to 0} (\frac{x}{\tan x}) \times \lim_{x \to 0} (- 2 \sqrt{x})

= 1 \times 0 = 0 .

d) \underset{x \to + \infty}{l i m} \frac{e^{2 x + 1} - e^{x}}{x^{2} - x - 1}

= \lim_{x \to \infty} e^{x} (\frac{e \times e^{x} - 1}{x^{2} - x + 1})

= \infty .

Commented by Rio Michael last updated on 13/Nov/19

t h a n k s s i r