Question

Help me With thiss PLEASE!!!
lim 1-2x^2-2cosx+cos^2x/x^1/2
approaches x -> 0

Answer 1

\[\lim_{x \rightarrow 0}\frac{1-2x^2-2\cos(x)+\cos^2(x)}{x^\frac{1}{2}}\]

Answer 2

this might be helpful
\[\cos^2(x)-2\cos(x)+1=(\cos(x)-1)^2 \]

Answer 3

\[\lim_{x \rightarrow 0}\frac{(\cos(x)-1)^2-2x^2}{x^\frac{1}{2}} \\ \lim_{x \rightarrow 0}\frac{x^\frac{3}{2}}{x^\frac{3}{2}}\frac{(\cos(x)-1)^2-2x^2}{x^\frac{1}{2}} \\ \lim_{x \rightarrow 0}\frac{ x^\frac{3}{2}(\cos(x)-1)^2-x^\frac{3}{2}2x^2}{x^2} \\ \lim_{x \rightarrow 0} x^\frac{3}{2} \frac{(\cos(x)-1)^2}{x^2}-\lim_{x \rightarrow 0}\frac{x^\frac{3}{2} 2x^2}{x^2} \\ \lim_{x \rightarrow 0}x^\frac{3}{2} (\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x})^2-\lim_{x \rightarrow 0}2x^\frac{3}{2}\]

Answer 4

and I kinda think they mean 0^+ since x^(1/2) doesn't actually exist for the real numbers when x<0

Answer 5

must be x>0

Answer 6

In my answer sheet appear -2 D: Im confused S:

Answer 7

I wrote down the function you wanted right?

Answer 8

I don't see how they get -2

Answer 9

\[\lim_{x \rightarrow 0}\frac{(\cos(x)-1)^2-2x^2}{x^\frac{1}{2}} \\ \lim_{x \rightarrow 0}\frac{x^\frac{3}{2}}{x^\frac{3}{2}}\frac{(\cos(x)-1)^2-2x^2}{x^\frac{1}{2}} \\ \lim_{x \rightarrow 0}\frac{ x^\frac{3}{2}(\cos(x)-1)^2-x^\frac{3}{2}2x^2}{x^2} \\ \lim_{x \rightarrow 0} x^\frac{3}{2} \frac{(\cos(x)-1)^2}{x^2}-\lim_{x \rightarrow 0}\frac{x^\frac{3}{2} 2x^2}{x^2} \\ \lim_{x \rightarrow 0}x^\frac{3}{2} (\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x})^2-\lim_{x \rightarrow 0}2x^\frac{3}{2}=0(0)^2-2(0) \\ =0-0=0\]

Answer 10

assuming we are looking at the right limit of course

Answer 11

wait damn it I made a mistake the denominator is x^2 !! not x^1/2 D:

Answer 12

\[\lim_{x \rightarrow 0}\frac{1-2x^2-2\cos(x)+\cos^2(x)}{x^2} \\ \lim_{x \rightarrow 0}\frac{(\cos(x)-1)^2-2x^2}{x^2}\]
separate the fraction
it isn't too much work after that

Answer 13

hahahaha Oh what a wonderful math night S: I almost got insane

Answer 14

math can make one insane

Answer 15

just check out Einstein's hair

Answer 16

anyways I must go
you have fun

Answer 17

THANKSS :D jajaja