Question

lim x → 1 2/1-x^2 + 1/x-1

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{2}{1 - x^{2} } + \dfrac{1}{x - 1} \bigg)$

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{2}{1 - x^{2} } - \dfrac{1}{1 - x} \bigg)$

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{2 - 1(1 + x)}{(1 - x)(1 + x)} \bigg)$

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{2 - 1 - x}{(1 - x)(1 + x)} \bigg)$

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{1 - x}{(1 - x)(1 + x)} \bigg)$

As we know that,

First we put the value of x = 0 in equation and check their indeterminant form.

$\sf \implies \lim_{x \to1} \bigg(\dfrac{1 - 1}{(1 - 1)(1 + 1)} \bigg)$

$\sf \implies \lim_{x \to 1} \dfrac{0}{0}$

As we can see,

This is the 0/0 form of indeterminant.

We can just simplify the equation, we get.

$\sf \implies \lim_{x \to 1} \bigg(\dfrac{1}{1 + x} \bigg)$

Put the value of x = 1 in equation, we get.

$\sf \implies \lim_{x \to 1} = \dfrac{1}{2}$

                                                                                                                       

MORE INFORMATION.

Some standard expansions.

(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! + ..

Answer 2

Answer:

here is answer dear friend