Question

Limit: infinity by infinity form​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2x - 1)^{6}(3x - 1)^{4} }{(2x + 1)^{10} }$

As we know that,

Firstly, we put the value of x = ∞ and check the indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{[2(\infty) - 1]^{6} [3(\infty) - 1]^{4} }{[2(\infty) + 1]^{10} }$

As we can see that,

It is in the form of ∞/∞ indeterminant, we get.

Multiply and divide numerator and denominator by x, we get.

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{\bigg( 2 - \dfrac{1}{x}\bigg)^{6} \bigg( 3 - \dfrac{1}{x} \bigg)^{4} }{\bigg(2 +\dfrac{1}{x} \bigg)^{10} }$

Put the value of x = ∞ in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{\bigg( 2 - \dfrac{1}{\infty}\bigg)^{6} \bigg( 3 - \dfrac{1}{\infty} \bigg)^{4} }{\bigg(2 +\dfrac{1}{\infty} \bigg)^{10} }$

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2)^{6} (3)^{4} }{(2)^{10} } = \dfrac{(3)^{4} }{(2)^{4} } = \dfrac{81}{16}$

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2x - 1)^{6}(3x - 1)^{4} }{(2x + 1)^{10} } = \dfrac{81}{16}$

                                                                                                                     

MORE INFORMATION.

(1) = eˣ = 1 + x + x²/2! + x³/3! + . . . . .

(2) = e⁻ˣ = 1 - x + x²/2! - x³/3! + . . . . .

(3) = ㏒(1 + x) = x - x²/2 + x³/3 - . . . . .

(4) = ㏒(1 - x) = - x - x²/2 - x³/3 - . . . . .

(5) = aˣ = 1 + (x ㏒ a) + (x ㏒ a)²/2! + (x ㏒ a)³/3! + . . . . .

(6) = sin x = x - x³/3! + x⁵/5! - . . . . .

(7) = cos x = 1 - x²/2! + x⁴/4! - . . . . .

(8) = tan x = x + x³/3 + 2x⁵/15 + . . . . .

Answer 2

Answer:

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2x - 1)^{6}(3x - 1)^{4} }{(2x + 1)^{10}}$

As we know that,

Firstly, we put the value of x = ∞ and check the indeterminant form, we get.

$\boxed{ \sf \implies \displaystyle \lim_{x \to \infty} \dfrac{[2(\infty) - 1]^{6} [3(\infty) - 1]^{4} }{[2(\infty) + 1]^{10} }}$

As we can see that,

It is in the form of ∞/∞ indeterminant, we get.

Multiply and divide numerator and denominator by x, we get.

$\boxed{\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{\bigg( 2 - \dfrac{1}{x}\bigg)^{6} \bigg( 3 - \dfrac{1}{x} \bigg)^{4} }{\bigg(2 +\dfrac{1}{x} \bigg)^{10} }}$

Put the value of x = ∞ in the equation, we get.

$\boxed{\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{\bigg( 2 - \dfrac{1}{\infty}\bigg)^{6} \bigg( 3 - \dfrac{1}{\infty} \bigg)^{4} }{\bigg(2 +\dfrac{1}{\infty} \bigg)^{10}}}$

$\boxed{ \sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2)^{6} (3)^{4} }{(2)^{10} } = \dfrac{(3)^{4} }{(2)^{4} } = \dfrac{81}{16}}$

81/16

Hence vertified

$\boxed{\sf \implies \displaystyle \lim_{x \to \infty} \dfrac{(2x - 1)^{6}(3x - 1)^{4} }{(2x + 1)^{10} } = \dfrac{81}{16}}$