Question

Evaluate : lim x tends to zero √(1+3x) - √(1-3x) whole divided by x

Please answer fast​

Answer 1

hope it's still not too late

Answer 2

$\red{ Value \: of \: \lim_{x\to 0} \: \frac{\sqrt{(1+3x)} - \sqrt{(1-3x)}}{x} }$

$= \lim_{x\to 0} \: \frac{[\sqrt{(1+3x)} - \sqrt{(1-3x)}][\sqrt{(1+3x)} + \sqrt{(1-3x)}]}{x[\sqrt{(1+3x)} +\sqrt{(1-3x)}] }$

$= \lim_{x\to 0} \: \frac{\sqrt{(1+3x)}^{2} - \sqrt{(1-3x)}^{2}}{x[\sqrt{(1+3x)} +\sqrt{(1-3x)}]}$

$= \lim_{x\to 0} \: \frac{(1+3x)- (1-3x)}{x[\sqrt{(1+3x)} +\sqrt{(1-3x)}]}$

$= \lim_{x\to 0} \: \frac{1+3x - 1+3x}{x[\sqrt{(1+3x)} +\sqrt{(1-3x)}]}$

$= \lim_{x\to 0 } \: \frac{6x}{x[\sqrt{(1+3x)} +\sqrt{(1-3x)}]}$

$= \lim_{x\to 0 } \: \frac{6}{\sqrt{(1+3x)} +\sqrt{(1-3x)} }$

$= \frac{ 3}{\sqrt{(1+3\times 0)} + \sqrt{(1-3\times 0)}}$

$= \frac{6}{1+1} \\= \frac{6}{2}\\= 3$

Therefore.,

$\red{ Value \: of \: \lim_{x\to 0} \: \frac{\sqrt{(1+3x)} - \sqrt{(1-3x)}}{x} } \green {= 3 }$

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