2 years ago

Passing through (5, -4) and parallel to the line whose equation is y=-2x+3; point-slope form

Mathematics

1 answer:

Marina CMI [18]2 years ago

8 0

The answer is y=-2x+6

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

Triss [41]

Answer:

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

Step-by-step explanation:

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