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aleksley [76]
3 years ago
7

Calculate the following limit:

Mathematics
1 answer:
aleksklad [387]3 years ago
8 0
\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\
=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1


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The square of a number is 12 less than 7 times the number. What is the number?
hjlf

<u><em>Answer:</em></u>

The number is either 3 or 4


<u><em>Explanation:</em></u>

Assume that the number we are looking for is x

The square of the number would be x²

7 times the number would be 7x


<u>We are given that:</u>

square of the number (x²) is 12 less than 7 times the number (7x)


<u>Translating this into an equation, we would have:</u>

x² = 7x - 12


<u>Rearranging the quadratic equation:</u>

x² - 7x + 12 = 0


<u>Now, we factor the equation:</u>

(x-4)(x-3) = 0


<u>This means that:</u>

either x-4 = 0 ..................> x = 4

or x-3 = 0 ........................> x = 3


<u>Let's check if any of the solutions is an extrovert:</u>

For x = 3 .............> (3)² = 7(3) - 12 .........> 9 = 21 - 12 .......> 9 = 9 .......> not extrovert

For x = 4 .............> (4)² = 7(4) - 12 .........> 16 = 28 - 12 .....> 16 = 16 ....> not extrovert


Since we have no restrictions on the number, therefore, both solutions would be accepted


Hope this helps :)


7 0
3 years ago
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