Question

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

Answer 1

Answer:

x = 0 + 2πk

√2 / 144

Step-by-step 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.

$\lim_{x \to 3} \frac{\sqrt{5+\sqrt{2x+3} }\ -\ 2\sqrt{2} }{x^{2}-9} \times \frac{\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2} }{\sqrt{5+\sqrt{2x+3} }\ +\ 2\sqrt{2}}$

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

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

Multiply by the conjugate of the new numerator.

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

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

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

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

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

$\frac{2}{(6)(4\sqrt{2})(6)}$

$\frac{\sqrt{2}}{144}$

Answer 2

$\bold{\text{Answer:}\quad \dfrac{\sqrt2}{144}}$

Step-by-step 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.

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