Question

Evaluate
$\rm \displaystyle \lim_{x\to 0}\ \frac{\sqrt{1+x+x^{2}}-1}{x}$

Answer 1

Solution :

$\rm \displaystyle \lim_{x\to 0}\ \frac{\sqrt{1+x+x^{2}}-1}{x}$

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[√(1+x+x²) -1][√(1+x+x²)+1]/{x[(√(1+x+x²)+1]}

= [(√1+x+x²)² - 1²]/[x(√1+x+x² + 1 )]

= ( 1+x+x²-1 )/[x(√1+x+x² + 1 )]

= [x(1 + x )]/[x(√1+x+x² + 1 )]

= ( 1 + x )/( √1+x+x² + 1 )

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$\rm \displaystyle \lim_{x\to 0}\ \frac{(1+x)}{\sqrt{1+x+x^{2}}+1}$

= ( 1 + 0 )/[ √(1+0+0) + 1 ]

= 1/( 1 + 1 )

= 1/2

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