Question

limits tends to 0 √1+x+x square - √x+1 / 2x square​

Answer 1

Answer::

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$\huge{\underline{\underline{{\pink{\sf \:Required \: Answer }}}}}$

$\pink\bigstar\sf{lim_x \to 0 \: \: \dfrac{\sqrt(1+x+x^2)-(\sqrt( x+1)}{2x^2}}$

$\: \: \: \: \: \implies \displaystyle \boxed{\rm{\clubsuit\ rationalize \: the \: equation\clubsuit}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \dfrac{(\sqrt(1+x+x^2))-(\sqrt (x+1))}{2x^2} \times \dfrac{(\sqrt( 1+x+x^2)+(\sqrt( x+1)}{(\sqrt( 1+x+x^2)+(\sqrt (x+1)}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \dfrac{(1+x+x^2)-(x+1)}{x^2 \times (\sqrt (1+x+x^2)+(\sqrt( x+1)}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \dfrac{(1+x+x^2-x-1)}{x^2 \times(\sqrt (1+x+x^2)+(\sqrt (x+1)}}$

$\dashrightarrow\sf{lim_x \to 0\: \: \: \dfrac{x^2}{x^2 \times(\sqrt (1+x+x^2)+(\sqrt( x+1)}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \cancel\dfrac{x^2}{x^2} \times\dfrac{1}{(\sqrt(1+x+x^2)+(\sqrt( x+1)}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \dfrac{1}{(\sqrt(1+x+x^2)+(\sqrt (x+1)}}$

$\: \: \: \: \: \: \star \sf{ \: put \: the \: value \: 0 \: in \: the \: equation \: of \: x}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \: \dfrac{1}{(\sqrt(1)+0+(0)^2)+(\sqrt(0)+1)}}$

$\dashrightarrow\sf{lim_x \to 0\: \: \: \dfrac{1}{1+1}}$

$\dashrightarrow\sf{lim_x \to 0 \: \: \implies\dfrac{1}{2}}$

$\: \: \: \: \: \: \blue \bigstar \boxed{ \sf \pink{ \: the \: answer \: is \: \dfrac{1}{2} }}$

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