Calculate-these-limits-Please-sirs-detail-

Question Number 87168 by mathocean1 last updated on 03/Apr/20

C a l c u l a t e t h e s e l i m i t s :

P l e a s e s i r s d e t a i l

Commented by john santu last updated on 03/Apr/20

a) change x \Rightarrow - t

\lim_{t \to \infty} \frac{- 3 t - 1 - \cos t}{- t} = 3

c) \lim_{x \to 4} \frac{\frac{1}{\sqrt{2 x + 1}}}{1 - \frac{3}{2 \sqrt{3 x + 4}}} = \frac{\frac{1}{3}}{1 - \frac{3}{8}} = \frac{\frac{1}{3}}{\frac{5}{8}} = \frac{8}{15}

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

Commented by mathocean1 last updated on 03/Apr/20

Commented by john santu last updated on 03/Apr/20

\lim_{n \to \infty} or \lim_{x \to \infty}

Commented by mathocean1 last updated on 03/Apr/20

l i \underset{x \to \infty}{m}

Commented by TANMAY PANACEA. last updated on 03/Apr/20

c) \lim_{x \to 4} \frac{\sqrt{2 x + 1} - 3}{x - \sqrt{3 x + 4}}

= \lim_{x \to 4} \frac{2 x + 1 - 9}{x^{2} - 3 x - 4} \times \frac{x + \sqrt{3 x + 4}}{\sqrt{2 x + 1} + 3}

= \lim_{x \to 4} \frac{2 (x - 4)}{(x - 4) (x + 1)} \times \frac{x + \sqrt{3 x + 4}}{\sqrt{2 x + 1} + 3}

= \frac{2}{5} \times \frac{4 + 4}{3 + 3} = \frac{2}{5} \times \frac{4}{3} = \frac{8}{15}

Commented by TANMAY PANACEA. last updated on 03/Apr/20

d) \lim_{x \to \infty} \frac{\sqrt{3 x^{2} + 1} - 2 x}{x}

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

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

Commented by mathmax by abdo last updated on 03/Apr/20

a) w e h a v e {l i m}_{x \to - \infty} \frac{3 x + 1 - c o s x}{x}

= {l i m}_{x \to - \infty} \frac{x (3 + \frac{1}{x} - \frac{c o s x}{x})}{x} = {l i m}_{x \to - \infty} 3 + \frac{1}{x} - \frac{c o s x}{x} = 3

Commented by mathmax by abdo last updated on 03/Apr/20

c) {l i m}_{x \to 4} \frac{\sqrt{2 x + 1} - 3}{x - \sqrt{3 x + 4}} = {l i m}_{x \to 4} \frac{2 x + 1 - 9}{\sqrt{2 x + 1} + 3} \times \frac{x + \sqrt{3 x + 4}}{x^{2} - 3 x - 4}

= {l i m}_{x \to 4} \frac{x + \sqrt{3 x + 4}}{\sqrt{2 x + 1} + 3} \times \frac{2 x - 8}{(x^{2} - 3 x - 4)} = \frac{4}{3} {l i m}_{x \to 4} \frac{2 (x - 4)}{(x - 4) (x + 1)}

= \frac{4}{3} {l i m}_{x \to 4} \frac{2}{x + 1} = \frac{4}{3} \times \frac{2}{5} = \frac{8}{15}

Commented by mathmax by abdo last updated on 03/Apr/20

d) f (x) = \frac{\sqrt{3 x^{2} + 1} - 2 x}{x} w e h a v e f o r x > 0

\sqrt{3 x^{2} + 1} = x \sqrt{3} \sqrt{1 + \frac{1}{3 x^{2}}} \sim x \sqrt{3} {1 + \frac{1}{6 x^{2}}} \Rightarrow f (x) \sim \frac{x \sqrt{3} + \frac{\sqrt{3}}{6 x} - 2 x}{x}

= \frac{\sqrt{3} + \frac{\sqrt{3}}{6 x^{2}} - 2}{1} \to \sqrt{3} - 2 \Rightarrow {l i m}_{x \to + \infty} f (x) = \sqrt{3} - 2

Commented by mathmax by abdo last updated on 03/Apr/20

b) l e t f (x) = \frac{x^{2} + 1}{x [x]} c h a n g e m e n t x = - t g i v e

f (x) = \frac{t^{2} + 1}{- t [- t]} (x \to - \infty \Rightarrow t \to + \infty) a n d w e h a v e

[- t] = α - [t] w i t h α = 0 o r - 1 \Rightarrow [- t] \sim - [t] \sim - t \Rightarrow

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

Commented by john santu last updated on 03/Apr/20

b) what define E (x) ?

Commented by john santu last updated on 03/Apr/20

why E (x) = [x] ?

E (x) = μ ?

Commented by mathmax by abdo last updated on 03/Apr/20

E (x) = x a t \infty

Commented by mathmax by abdo last updated on 03/Apr/20

x = [x] + r w i t h 0 ⩽ r < 1 \Rightarrow \frac{x}{[x]} = 1 + \frac{r}{[x]} \Rightarrow \frac{x}{[x]} \sim 1 (x \to + \infty)

Commented by mathocean1 last updated on 03/Apr/20

E (x) ⩽ x ⩽ E (x) + 1