Math Question and Answers


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 80585 by ahmadshahhimat775@gmail.com last updated on 04/Feb/20

Commented by kaivan.ahmadi last updated on 04/Feb/20

$2 .$ ${l i m}_{x \to \frac{π}{6}} \frac{2 s i n 2 x + c o s x}{2 s i n 2 x - 3 c o s x} = \frac{\sqrt{3} + \frac{\sqrt{3}}{2}}{\sqrt{3} - \frac{3 \sqrt{3}}{2}} = \frac{\frac{3 \sqrt{3}}{2}}{\frac{- \sqrt{3}}{2}} = - 3$
Commented by kaivan.ahmadi last updated on 04/Feb/20

$5 .$ ${l i m}_{x \to \frac{π}{4}} (t g x - 1) t g 2 x = {l i m}_{x \to \frac{π}{4}} \frac{t g x - 1}{c o t 2 x} =$ ${l i m}_{x \to \frac{π}{4}} \frac{1 + {t g}^{2} x}{- 2 (1 + {c o t}^{2} 2 x)} = \frac{1 + 1}{- 2 (1 + 0)} = - 1$
Commented by kaivan.ahmadi last updated on 04/Feb/20

$6 .$ ${l i m}_{x \to e} \frac{l n x + 1 - 1}{\frac{2}{x}} = \frac{1 + 1 - 1}{\frac{2}{e}} = \frac{e}{2}$
Commented by mathmax by abdo last updated on 04/Feb/20
$1) let A_n =((2^n (n!))/n^n ) we have n!∼ n^n e^(−n) (√(2πn)) ⇒ A_n ∼((2^n .n^n e^(−n) (√(2π))n)/n^n ) =((2/e))^n (√(2πn)) =(√(2π))e^(nln((2/e))) ×e^((1/2)ln(n)) =(√(2π))×e^(nln((2/e))+ln((√n))) but we hsve nln((2/e))+ln((√n)) =nln(2)−n +(1/2)ln(n) =n{ln(2)−1+((ln(n))/(2n))}→+∞ ⇒ lim_(n→+∞) A_n =+∞$
$1) l e t A_{n} = \frac{2^{n} (n!)}{n^{n}} w e h a v e n! \sim n^{n} e^{- n} \sqrt{2 π n} \Rightarrow$ $A_{n} \sim \frac{2^{n} . n^{n} e^{- n} \sqrt{2 π} n}{n^{n}} = {(\frac{2}{e})}^{n} \sqrt{2 π n} = \sqrt{2 π} e^{n l n (\frac{2}{e})} \times e^{\frac{1}{2} l n (n)}$ $= \sqrt{2 π} \times e^{n l n (\frac{2}{e}) + l n (\sqrt{n})} b u t w e h s v e$ $n l n (\frac{2}{e}) + l n (\sqrt{n}) = n l n (2) - n + \frac{1}{2} l n (n)$ $= n {l n (2) - 1 + \frac{l n (n)}{2 n}} \to + \infty \Rightarrow {l i m}_{n \to + \infty} A_{n} = + \infty$
Commented by mathmax by abdo last updated on 04/Feb/20
$forgive ln(2)−1<0 ⇒n{ln(2)−1+((ln(n))/(2n))}→−∞ ⇒ lim_(n→∞) A_n =0$
$f o r g i v e l n (2) - 1 < 0 \Rightarrow n {l n (2) - 1 + \frac{l n (n)}{2 n}} \to - \infty \Rightarrow$ ${l i m}_{n \to \infty} A_{n} = 0$
Commented by mathmax by abdo last updated on 04/Feb/20

$2) c h a n g e m e n t s i n x = t g i v e$ ${l i m}_{x \to \frac{π}{6}} \frac{2 {s i n}^{2} x + s i n x - 1}{2 {s i n}^{2} x - 3 s i n x + 1} = {l i m}_{t \to \frac{1}{2}} \frac{2 t^{2} + t - 1}{2 t^{2} - 3 t + 1}$ $2 t^{2} + t - 1 = 2 (t - \frac{1}{2}) (t - a) = (2 t - 1) (t - a) \Rightarrow a = - 1 \Rightarrow$ $2 t^{2} + t - 1 = (2 t - 1) (t + 1)$ $2 t^{2} - 3 t + 1 = 2 (t - \frac{1}{2}) (t - b) = (2 t - 1) (t - b) \Rightarrow b = 1 \Rightarrow$ $2 t^{2} - 3 t + 1 = (2 t - 1) (t - 1) \Rightarrow$ ${l i m}_{t \to \frac{1}{2}} (. . .) = {l i m}_{t \to \frac{1}{2}} \frac{(2 t - 1) (t + 1)}{(2 t - 1) (t - 1)} = {l i m}_{t \to \frac{1}{2}} \frac{t + 1}{t - 1} = \frac{\frac{3}{2}}{- \frac{1}{2}} = - 3$
Commented by mathmax by abdo last updated on 04/Feb/20
$let X_n =((n!)/4^n ) we have n!∼n^(n ) e^(−n) (√(2πn)) ⇒ X_n ∼ (n^n /((4e)^n ))(√(2π))n =e^(nln(n)) e^(−nln(4e)) (√(2π))e^((1/2)ln(n)) =e^(nln(n)−nln(4e)+((ln(n))/2)) =e^(n{ln(n)−ln(4e)+((ln(n))/(2n))}) →+∞$
$l e t X_{n} = \frac{n!}{4^{n}} w e h a v e n! \sim n^{n} e^{- n} \sqrt{2 π n} \Rightarrow$ $X_{n} \sim \frac{n^{n}}{{(4 e)}^{n}} \sqrt{2 π} n = e^{n l n (n)} e^{- n l n (4 e)} \sqrt{2 π} e^{\frac{1}{2} l n (n)}$ $= e^{n l n (n) - n l n (4 e) + \frac{l n (n)}{2}} = e^{n {l n (n) - l n (4 e) + \frac{l n (n)}{2 n}}} \to + \infty$
Commented by mathmax by abdo last updated on 04/Feb/20

$5) l e t f (x) = {(t a n x)}^{t a n (2 x)} \Rightarrow f (x) = e^{t a n (2 x) l n (t a n x)}$ $c h a n g e m e n t x = \frac{π}{4} + t g i v e x \to \frac{π}{4} \Leftrightarrow t \to 0$ $f (x) = g (t) = e^{t a n (2 (\frac{π}{4} + t)) l n (t a n (\frac{π}{4} + t))}$ $= e^{t a n (2 t + \frac{π}{2}) l n (t a n (\frac{π}{4} + t))} w e h a v e$ $t a n (2 t + \frac{π}{2}) = \frac{s i n (2 t + \frac{π}{2})}{c o s (2 t + \frac{π}{2})} = - \frac{1}{t a n (2 t)} \Rightarrow$ $t a n (\frac{π}{4} + t) = \frac{1 + t a n (t)}{1 - t a n (t)} \Rightarrow l n (t a n (\frac{π}{4} + t)) = l n (1 + t a n t) - l n (1 - t a n t)$ $\Rightarrow t a n (2 t + \frac{π}{2}) l n (t a n (\frac{π}{4} + t)) \sim \frac{- 1}{2 t} \times (t - (- t)) = - 1 \Rightarrow$ ${l i m}_{t \to 0} g (t) = \frac{1}{e} = {l i m}_{x \to \frac{π}{4}}$
Commented by mathmax by abdo last updated on 04/Feb/20

${l i m}_{x \to \frac{π}{4}} f (x) = \frac{1}{e}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum