displaystyle lim x arrow ∞ ((2 x+1)40(4 x-1)5/(2 x+3)45) is equal to

Question

displaystyle lim x arrow ∞ ((2 x+1)40(4 x-1)5/(2 x+3)45) is equal to (A) 16 (B) 24 (C) 32 (D) 8

Tardigrade Admin · Accepted Answer

displaystyle lim x arrow ∞ ((2 x+1)40(4 x-1)5/(2 x+3)45) = displaystyle lim x arrow ∞ ((2+(1/x))40(4-(1/x))5/(2+(3)x)45) (Dividing numerator and denominator by x45 ) =(240 45/245) =25=32

Q. $x \to \infty lim \frac{( 2 x + 1 ) ^{40} ( 4 x - 1 ) ^{5}}{( 2 x + 3 ) ^{45}}$ is equal to

16

24

32

8

$x \to \infty lim \frac{( 2 x + 1 ) ^{40} ( 4 x - 1 ) ^{5}}{( 2 x + 3 ) ^{45}}$
$= x \to \infty lim \frac{( 2 + \frac{1}{x} ) ^{40} ( 4 - \frac{1}{x} ) ^{5}}{( 2 + \frac{3}{x} ) ^{45}}$
(Dividing numerator and denominator by $x^{45}$ )
$= \frac{2 ^{40} 4 ^{5}}{2 ^{45}}$
$= 2^{5} = 32$

Q. x→∞lim​(2x+3)45(2x+1)40(4x−1)5​ is equal to

16

24

32

8

x→∞lim​(2x+3)45(2x+1)40(4x−1)5​ =x→∞lim​(2+x3​)45(2+x1​)40(4−x1​)5​ (Dividing numerator and denominator by x45 ) =24524045​ =25=32

Q. $x \to \infty lim \frac{( 2 x + 1 ) ^{40} ( 4 x - 1 ) ^{5}}{( 2 x + 3 ) ^{45}}$ is equal to

$x \to \infty lim \frac{( 2 x + 1 ) ^{40} ( 4 x - 1 ) ^{5}}{( 2 x + 3 ) ^{45}}$
$= x \to \infty lim \frac{( 2 + \frac{1}{x} ) ^{40} ( 4 - \frac{1}{x} ) ^{5}}{( 2 + \frac{3}{x} ) ^{45}}$
(Dividing numerator and denominator by $x^{45}$ )
$= \frac{2 ^{40} 4 ^{5}}{2 ^{45}}$
$= 2^{5} = 32$