Limit Problem- Please help- It's easy but my TI broke:(

[mathqueen88]

f(x) = (3)(cos2x-1) / 4x^2


x->0


Now basically the 2 between cos - x is a square , and so is the 2 in the denominator .


The question is, what is the Right hand Limit and Left hand limit ( I got .347 and 0 ) respectively,

And what is the normal limit ? ( I got DNE)

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \frac{3cos(2x-1)}{4x^2}$

$\displaystyle \lim_{x\to0}f(x) \ = \ \frac{3cos(-1)}{0} \ = \ \infty.$

 

[mathqueen88]

the Cos is squared, that is not a 2x but instead a cos^2(x) so thus cos(x)^2

and I asked for left and right hand limit too

 

[BigGlenntheHeavy]

$\displaystyle Hey \ mathqueen88, \ I'm \ not \ a \ Swami, \ perhaps \ others \ on \ this \ board \ are.$

$\displaystyle cos(2x) \ \ne \ cos^2(x) \ and \ cos^2(x) \ = \ [cos(x)]^2 \ \ne \ cos(x)^2.$

$\displaystyle Learn \ your \ grouping \ symbols \ as, \ at \ present, \ they're \ atrocious.$

 

[Deleted member 4993]

f(x) = (3)(cos2x-1) / 4x^2


x->0


Now basically the 2 between cos - x is a square , and so is the 2 in the denominator .


The question is, what is the Right hand Limit and Left hand limit ( I got .347 and 0 ) respectively,

And what is the normal limit ? ( I got DNE)

After all the explanation:

$\displaystyle \lim_{x\to 0}\frac{3[cos^2(x) - 1]}{4x^2}$

Does your problem look like the above?

 

[BigGlenntheHeavy]

$\displaystyle An \ another \ thing, \ if \ your \ problem \ is \ what \ Subhotosh \ Khan \ has \ quoted \ above,$

$\displaystyle then \ the \ limit \ as \ x \ approaches \ 0, \ 0^-, \ and \ 0^+ \ is \ -3/4, \ not \ the \ crap \ that \ you \ have$

$\displaystyle expounded.$

 

[galactus]

$\displaystyle \lim_{x\to 0}\frac{3(cos^{2}(x)-1)}{4x^{2}}$

Note that $\displaystyle cos^{2}(x)-1=-sin^{2}(x)$

So, we get:

$\displaystyle \frac{-3}{4}\lim_{x\to 0}\frac{sin^{2}(x)}{x^{2}}$

$\displaystyle \frac{-3}{4}\lim_{x\to 0}\left(\frac{sin(x)}{x}\right)^{2}$

Remember what $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}$ equals?.

See the limit now?. Since the limit exists, the right limit and left limit are the same.

By the way, use the ^ key for powers. Writing cos2x when you mean $\displaystyle cos^{2}(x)$ is going to confuse anyone.

Afterall, that is what the SHIFT 6 key is for.