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Hunter-Best [27]
3 years ago
7

Find the limit if it exists lim x→0 sqrtx+7-sqrt7 over x

Mathematics
1 answer:
Triss [41]3 years ago
6 0

Answer:

\frac{1}{ 2\sqrt{7} }

Step-by-step explanation:

\lim_{x\to 0}  \frac{ \sqrt{x + 7}  -  \sqrt{7} }{x}  \\  \\  = \lim_{x\to 0}  \frac{( \sqrt{x + 7}  -  \sqrt{7}) }{x}  \times  \frac{( \sqrt{x + 7}   +   \sqrt{7}) }{( \sqrt{x + 7}   +  \sqrt{7}) }  \\  \\   = \lim_{x\to 0}  \frac{( \sqrt{x + 7} )^{2}  -  (\sqrt{7})^{2}  }{x( \sqrt{x + 7}   +  \sqrt{7})}  \\  \\   = \lim_{x\to 0}  \frac{( {x + 7}  -  {7}) }{x( \sqrt{x + 7}   +  \sqrt{7})}   \\  \\ = \lim_{x\to 0}  \frac{ {\cancel x}}{\cancel x( \sqrt{x + 7}   +  \sqrt{7})} \\  \\ = \lim_{x\to 0}  \frac{ {1}}{\sqrt{x + 7}   +  \sqrt{7}}  \\  \\  =  \frac{1}{ \sqrt{0 + 7} +  \sqrt{7}  } \\  \\  =  \frac{1}{ \sqrt{7} +  \sqrt{7}  }  \\  \\  =  \frac{1}{ 2\sqrt{7} }

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