$\lim_{x \to \infty}\frac{1-cos4x}{1-cos2x}= \lim_{x \to 0} \frac{(1-cos4x)'}{(1-cos2x)'}= \lim_{x \to 0}\frac{4sin4x}{2sin2x}= \lim_{x \to 0}\frac{2*2sin2xcos2x}{sin2x}\\ = \lim_{x \to 0}4cos2x\\ =4$

{L'Hospital's rule}

<em>I hope this helps you</em>

6 0

2 years ago

Where do the lines modeled by the equations y =-7.8x+1.4 and y -3.4x +1.2intersect

vaieri [72.5K]

Answer:

sinfcn

Step-by-step explanation:

nelLdng 5kaidog tnn tee i k she nf

7 0

3 years ago

-4t > -20 step by step please

kenny6666 [7]

-4t > -20
To solve for t, you have to divide -4 on both sides:
-4t/-4 > -20/-4
t > 5 its your answer.
Hope this helps!!!

6 0

4 years ago