Question

please help me to solving this question​

Answer 1

Solution!!

$\sf \displaystyle \lim_{\sf x\to 0 }\dfrac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$

Multiply the fraction.

$\sf \implies \displaystyle \lim_{\sf x\to 0}\dfrac{\sqrt{1+3x}-\sqrt{1-3x}}{x}\times \dfrac{\sqrt{1+3x}+\sqrt{1-3x}}{\sqrt{1+3x}+\sqrt{1-3x}}$

Multiply the numerators and denominators separately.

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

Use (a - b)(a + b) = a² - b² to simplify the product.

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

Add and subtract the like terms.

$\sf \implies \displaystyle \lim_{\sf x\to 0}\dfrac{6x}{x(\sqrt{1+3x}+\sqrt{1-3x})}$

Reduce the fraction with x.

$\sf \implies \displaystyle \lim_{\sf x\to 0}\dfrac{6}{\sqrt{1+3x}+\sqrt{1-3x}}$

Evaluate the limit.

$\sf \implies \dfrac{6}{\sqrt{1+3\times 0}+\sqrt{1-3\times 0}}$

Calculate the value.

$\sf \implies \dfrac{6}{\sqrt{1+0}+\sqrt{1-0}}$

$\sf \implies \dfrac{6}{1+1}$

$\sf \implies 3$

Option (a) is correct.

Answer 2

correct option is : (a) 3