Question

limit X tends to zero under root 1+x+xsquare -1 by x
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Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\displaystyle\lim_{x\to\,0}\;\dfrac{\sqrt{1+x+x^2}-1}{x}}$

$\underline{\textbf{To evaluate:}}$

$\mathsf{\displaystyle\lim_{x\to\,0}\;\dfrac{\sqrt{1+x+x^2}-1}{x}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\displaystyle\lim_{x\to\,0}\;\dfrac{\sqrt{1+x+x^2}-1}{x}}$

$\mathsf{=\dfrac{1-1}{0}=\dfrac{0}{0}\;form}$

$\textsf{Apply L Hopital's rule}$

$\mathsf{=\displaystyle\lim_{x\to\,0}\;\dfrac{\dfrac{1}{2\sqrt{1+x+x^2}}(1+2x)}{1}}$

$\mathsf{=\dfrac{\dfrac{1}{2\sqrt{1+0+0^2}}(1+2(0))}{1}}$

$\mathsf{=\dfrac{1}{2}}$

$\implies\boxed{\mathsf{\displaystyle\lim_{x\to\,0}\;\dfrac{\sqrt{1+x+x^2}-1}{x}=\dfrac{1}{2}}}$

$\underline{\textbf{Find more:}}$

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Answer 2

SOLUTION

TO DETERMINE

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

EVALUATION

Here the given limit is

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

We solve it as below

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

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{( \sqrt{1 + x + {x}^{2} } - 1)(\sqrt{1 + x + {x}^{2} } + 1) }{x(\sqrt{1 + x + {x}^{2} } + 1)} }$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{{( \sqrt{1 + x + {x}^{2} })}^{2} - {(1)}^{2} }{x(\sqrt{1 + x + {x}^{2} } + 1)} }$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{(1 + x + {x}^{2} - 1) }{x(\sqrt{1 + x + {x}^{2} } + 1)} }$

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

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

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

$\displaystyle \sf{ = \: \frac{( 1 + 0 ) }{(\sqrt{1 + 0 + {0}^{2} } + 1)} }$

$\displaystyle \sf{ = \: \frac{1}{(\sqrt{1 } + 1)} }$

$\displaystyle \sf{ = \: \frac{1}{(1 + 1)} }$

$\displaystyle \sf{ = \: \frac{1}{2} }$

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