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.

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.

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.

Multiply by the conjugate of the new numerator.

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1