Trigonometric Limits- Is there a simpler way?

[ku1005]

Hi, just wondering if anyone could see a quicker way to solve the following limit, because it seems really long to me? Or maybe im wrong. Thanks anyway for looking!! Appreciated.

lim x-> 0 for (cos 2x - 2cosx +1)/ xcos x

What i did was split the limit into 3 parts:

lim x-> 0 cos 2x/ xcos x

lim x-> 0 -2cosx/xcosx and

lim x-> 0 1/xcosx

then i went about solving each of the seperate limits and summing the total in the following way

cos 2x/xcosx = 1/ xcosx * cos2x/1
= 1/xcos x * sinx/sinx (ie 1)
= (sinx/x) * (1/ cos x)* (1/sinx)
= sinx/sin^2 (x)
= sinx/(1-cos^2(x))
= sinx/1 * 1/-cos^2(x)
= which therefore, takin all the values so far is equivalent to
= 1*1*1*0*(-1) = 0

-2cosx/xcosx = 1/xcosx * -2cosx/1 [note here we already know 1/xcos x =0] therfore naturally this limit must = 0

and then

1/xcosx = 0, from above we know this = 0

therefore, summing these parts to give total

0 + 0 + 0 = 0 lol

so yeah, just wondering if i am doing this correctly first, and possibly if there is a quicker method?? thanks again
cheers


[/tex]

 

[galactus]

Hi, just wondering if anyone could see a quicker way to solve the following limit, because it seems really long to me? Or maybe im wrong. Thanks anyway for looking!! Appreciated.

lim x-> 0 for (cos 2x - 2cosx +1)/ xcos x

What i did was split the limit into 3 parts:

lim x-> 0 cos 2x/ xcos x

lim x-> 0 -2cosx/xcosx and

lim x-> 0 1/xcosx

then i went about solving each of the seperate limits and summing the total in the following way

cos 2x/xcosx = 1/ xcosx * cos2x/1
= 1/xcos x * sinx/sinx (ie 1)
= (sinx/x) * (1/ cos x)* (1/sinx)
= sinx/sin^2 (x)
= sinx/(1-cos^2(x))
= sinx/1 * 1/-cos^2(x)
= which therefore, takin all the values so far is equivalent to
= 1*1*1*0*(-1) = 0

-2cosx/xcosx = 1/xcosx * -2cosx/1 [note here we already know \(\displaystyle \L\\\underbrace{1/(xcos x) =0}_{\text{Actually, this is not 0, it's undefined}}\)] therfore naturally this limit must = 0

and then

1/xcosx = 0, from above we know this = 0

therefore, summing these parts to give total

0 + 0 + 0 = 0 lol

so yeah, just wondering if i am doing this correctly first, and possibly if there is a quicker method?? thanks again
cheers


[/tex]

\(\displaystyle \L\\\lim_{x\to\0}\frac{cos(2x)-2cos(x)+1}{xcos(x)}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{2cos^{2}(x)-1-2cos(x)+1}{xcos(x)}\)

\(\displaystyle \L\\2\lim_{x\to\0}\frac{cos(x)-1}{x}\)

Can you finish?.

 

[ku1005]

sure can!

thanks very much....and sorry about long reply, iv just been away for a couple of days! but thanks again! that limit = 0