Question

Help me solve them plsss...both of the questions :)

AND
I'm not sure whether the answer I got for the limit question is correct but i ended with a 0.
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Answer 1

x = 0 + 2πk

√2 / 144

Explanation:

cos x + cos 2x + cos 3x + cos 4x + cos 5x = 5

Cosine has a maximum of 1, so the only way 5 cosine terms can add up to 5 is if each one equals 1.

cos x = 1 → x = 0 + 2πk

cos 2x = 1 → x = 0 + πk

cos 3x = 1 → x = 0 + ⅔πk

cos 4x = 1 → x = 0 + ½πk

cos 5x = 1 → x = 0 + ⅖πk

The least common multiple of 2, 1, ⅔, ½, and ⅖ is 2.

So the solution that satisfies all five is x = 0 + 2πk.

lim(x→3) [√(5 + √(2x + 3)) − 2√2] / (x² − 9)

To solve without L'Hopital's rule, multiply by the conjugate of the numerator.

`\lim_(x \to 3) \frac{\sqrt{5+√(2x+3) }\ -\ 2√(2) }{x^(2)-9} * \frac{\sqrt{5+√(2x+3) }\ +\ 2√(2) }{\sqrt{5+√(2x+3) }\ +\ 2√(2)}`

`\lim_(x \to 3) \frac{5+√(2x+3)\ -\ 8}{(x^(2)-9)(\sqrt{5+√(2x+3) }\ +\ 2√(2))}`

`\lim_(x \to 3) \frac{√(2x+3)\ -\ 3}{(x^(2)-9)(\sqrt{5+√(2x+3) }\ +\ 2√(2))}`

Multiply by the conjugate of the new numerator.

`\lim_(x \to 3) \frac{√(2x+3)\ -\ 3}{(x^(2)-9)(\sqrt{5+√(2x+3) }\ +\ 2√(2))} *(√(2x+3)\ +\ 3)/(√(2x+3)\ +\ 3)`

`\lim_(x \to 3) \frac{2x+3-9}{(x^(2)-9)(\sqrt{5+√(2x+3) }\ +\ 2√(2))(√(2x+3)\ +\ 3)}`

`\lim_(x \to 3) \frac{2x-6}{(x^(2)-9)(\sqrt{5+√(2x+3) }\ +\ 2√(2))(√(2x+3)\ +\ 3)}`

`\lim_(x \to 3) \frac{2(x-3)}{(x-3)(x+3)(\sqrt{5+√(2x+3) }\ +\ 2√(2))(√(2x+3)\ +\ 3)}`

`\lim_(x \to 3) \frac{2}{(x+3)(\sqrt{5+√(2x+3) }\ +\ 2√(2))(√(2x+3)\ +\ 3)}`

`(2)/((6)(4√(2))(6))`

`(√(2))/(144)`

Answer 2

`\bold{\text{Answer:}\quad (\sqrt2)/(144)}`

Explanation:

When you plug in 3 directly to the equation, you get 0/0

Since it is indeterminate, you have to use L'Hopital's Rule.

That means that you find the limit of the DERIVATIVE of the numerator and the DERIVATIVE of the denominator.

`(d)/(dx)(\sqrt{5+√(2x+3)}-2\sqrt2)\quad = \frac{1}{2\sqrt{5+√(2x+3)}(√(2x+3))}\\\\\\f'(3)\ \text{for the numerator}\ =(1)/(12\sqrt2)\\\\\\\\(d)/(dx)(x^2-9) = 2x\\\\\\f'(3) \text{for the denominator}\quad = 6\\\\\\\frac{f'(3)\ \text{numerator}}{f'(3)\ \text{denominator}}\quad = ((1)/(12\sqrt2))/(6)\quad = (1)/(72\sqrt2)\quad \rightarrow \large\boxed{(\sqrt2)/(144)}`