Question

limit x tends to 0 √4+x -³√8+3x/x

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \sf{ \lim_{x \to 0} \: \frac{ \sqrt{4 + x} - \sqrt[3]{8 + 3x} }{x}}$

EVALUATION

$\displaystyle \sf{ \lim_{x \to 0} \: \frac{ \sqrt{4 + x} - \sqrt[3]{8 + 3x} }{x}} \: \: \: \: \: \: \: \: \: \bigg( \frac{0}{0} \: \: form \bigg)$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ {(4 + x)}^{ \frac{1}{2} } - {(8 + 3x)}^{ \frac{1}{3} } }{x}} \: \: \: \: \: \: \: \: \bigg( \frac{0}{0} \: \: form \bigg)$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ \frac{1}{2} {(4 + x)}^{ - \frac{1}{2} } - 3 \times \frac{1}{3} \times {(8 + 3x)}^{ - \frac{2}{3} } }{1}}$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ \frac{1}{2} {(4 + x)}^{ - \frac{1}{2} } - {(8 + 3x)}^{ - \frac{2}{3} } }{1}}$

$\displaystyle \sf{ = \frac{ \frac{1}{2} {(4 + 0)}^{ - \frac{1}{2} } - {(8 + 3.0)}^{ - \frac{2}{3} } }{1}}$

$\displaystyle \sf{ = \frac{1}{2} {(4 )}^{ - \frac{1}{2} } - {(8 )}^{ - \frac{2}{3} } }$

$\displaystyle \sf{ = \frac{1}{2} {( {2}^{2} )}^{ - \frac{1}{2} } - {( {2}^{3} )}^{ - \frac{2}{3} } }$

$\displaystyle \sf{ = \frac{1}{ {2}^{2} } - { {2}^{ - 2} } }$

$\displaystyle \sf{ = \frac{1}{4} - \frac{1}{4} }$

$= 0$

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