Question

lim x_0 log(1+x)/3^x-1​

Answer 1

EXPLANATION.

Method = 1.

$\implies \displaystyle \lim_{x \to 0} \dfrac{log(1 + x)}{3^{x} - 1}$

As we know that,

Put the value of x = 0 in the equation and check their indeterminant form, we get.

$\implies \displaystyle \lim_{x \to 0} \dfrac{log(1 + 0)}{3^{0} - 1} \ = \dfrac{0}{0}$

As we can see that,

It is in 0/0 form of indeterminant, we get.

Using L-HOSPITAL'S rule in the equation, we get.

$\implies \displaystyle \lim_{x \to 0} \dfrac{\dfrac{d}{dx}(log(1 + x)) }{\dfrac{d}{dx}(3^{x} - 1)}$

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{1}{1 + x}\bigg) }{3^{x} . log_{e}3}$

Put the value of x = 0 in the equation, we get.

$\implies \displaystyle \lim_{x \to 0} \dfrac{1/1}{log_{e}3} \ = \dfrac{1}{log_{e}3} \ = log_{3}e$

Method = 2.

$\implies \displaystyle \lim_{x \to 0} \dfrac{log(1 + x)}{3^{x} - 1}$

As we know that,

Divide numerator and denominator by x, we get.

$\implies \displaystyle \lim_{x \to 0} \dfrac{\bigg(\dfrac{log(1 + x)}{x} \bigg)}{\bigg(\dfrac{3^{x} - 1 }{x} \bigg)}$

$\implies \displaystyle \lim_{x \to 0} \dfrac{log(1 + x)}{3^{x} - 1 } \ = \dfrac{1}{log_{e}3} \ = log_{3}e$

Option [C] is correct answer.

                                                                                                                       

MORE INFORMATION.

Limits using expansion.

(1) = ㏒(1 + x) = x - x²/2 + x³/3 - x⁴/4 + x⁵/5 . . . . .

(2) = ㏒(1 - x) = - x - x²/2 - x³/3 - x⁴/4 - x⁵/5. . . . .

(3) = eˣ  1 + x + x²/2! + x³/3! + x⁴/4! + . . . . .

(4) = e⁻ˣ = 1 - x + x²/2! - x³/3! + x⁴/4! . . . . .

(5) = aˣ = 1 + x(㏒ a) + x²/2! (㏒ a)² + . . . . .

(6) = sin(x) = x - x³/3! + x⁵/5! - x⁷/7! . . . . .

(7) = cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! . . . . .

(8) = tan(x) = x + x³/3 + 2x²/15 + . . . . .