Question

limit x--->0 ln(1+x)- sinx+x^2/2 /xtanxsinx​

Answer 1

EXPLANATION.

$\implies \displaystyle \lim_{x \to 0} \dfrac{ln(1 + x) - sin(x) + \dfrac{x^{2} }{2} }{(x)(tan x)(sinx)}$

As we know that,

Formula of some standard limits,

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{tan(x)}{x} \bigg) = 1.$

$\implies \displaystyle \lim_{x \to 0} \bigg(\frac{sin(x)}{x} \bigg) = 1.$

$\implies \displaystyle log(1 + x) = x - \dfrac{x^{2} }{2} + \dfrac{x^{3} }{3} - . . . . .$

$\implies \displaystyle sin(x) = x - \dfrac{x^{3} }{3!} + \dfrac{x^{5} }{5!} - . . . . .$

Using this expansions in this equation, we get.

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(x - \dfrac{x^{2} }{2} + \dfrac{x^{3} }{3} \bigg) - \bigg(x - \dfrac{x^{3} }{3!}\bigg) + \dfrac{x^{2} }{2} }{(x) \bigg(\dfrac{tan x}{x} \bigg) \bigg(\dfrac{sinx}{x}\bigg) (x^{2} ) }$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg( x - \dfrac{x^{2} }{2} + \dfrac{x^{3} }{3} - x + \dfrac{x^{3} }{3!} + \dfrac{x^{2} }{2} \bigg)}{x^{3} }$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{x^{3} }{3} + \dfrac{x^{3} }{3!} \bigg)}{x^{3} }$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{x^{3} }{3} + \dfrac{x^{3} }{6}\bigg) }{x^{3} }$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{2x^{3} + x^{3} }{6} \bigg)}{x^{3} }$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{3x^{3} }{6} \bigg)}{x^{3} } = \dfrac{3}{6} = \dfrac{1}{2}$

$\implies \displaystyle \lim_{x \to 0} \dfrac{ln(1 + x) - sin(x) + \dfrac{x^{2} }{2} }{(x)(tan x)(sinx)} = \dfrac{1}{2}$

                                                                                                                       

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 + . . . . .