bemath-lim-x-2-3x-1-4x-2-

Question Number 107028 by bemath last updated on 08/Aug/20

@ b e m a t h @

\lim_{x \to \infty} {(2 + 3 x)}^{\frac{1}{4 x - 2}}

Commented by kaivan.ahmadi last updated on 08/Aug/20

y = {(2 + 3 x)}^{\frac{1}{4 x - 2}} \Rightarrow l n y = \frac{l n (2 + 3 x)}{4 x - 2} \Rightarrow {l i m}_{x \to \infty} l n y =

{l i m}_{x \to \infty} \frac{l n (2 + 3 x)}{4 x - 2} \sim {l i m}_{x \to \infty} \frac{3}{4 (2 + 3 x)} = 0

\Rightarrow {l i m}_{x \to \infty} y = e^{0} = 1

Answered by Dwaipayan Shikari last updated on 08/Aug/20

y = \lim_{x \to \infty} {(2 + 3 x)}^{\frac{1}{4 x - 2}}

logy = \lim_{x \to \infty} \frac{\log (2 + 3 x)}{4 x - 2} = \lim_{x \to \infty} \frac{3}{4 (2 + 3 x)} = 0

y = e^{0} = 1

Answered by john santu last updated on 08/Aug/20

@ JS @

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

= {[\lim_{x \to \infty} {(1 + \frac{1}{(\frac{1}{3 x + 1})})}^{3 x + 1}]}^{\frac{1}{(3 x + 1) (4 x - 2)}}

= e^{\lim_{x \to \infty} (\frac{1}{(3 x + 1) (4 x - 2)})} = e^{0} = 1

Answered by mathocean1 last updated on 08/Aug/20

=> l e t 2 + 3 x = a .

\underset{x \to \infty}{l i m} a^{\frac{1}{4 (\infty) - 2}} = \underset{x \to \infty}{l i m} a^{0} = \underset{x \to \infty}{l i m} 1 = 1

“ {o c e a n}^{″}