1. ## Cosine Trig Limit

limx->0 cosx-1/2x

ok i kinda dont know where to start

2. ## Re: Cosine Trig Limit

Originally Posted by purplec16
limx->0 cosx-1/2x

ok i kinda dont know where to start
It should be limx->0 {cosx-1}/2x.

3. ## Re: Cosine Trig Limit

Originally Posted by purplec16
limx->0 cosx-1/2x

ok i kinda dont know where to start
edit: post put in a spoiler now it apparent what the OP means.

Spoiler:
What is that second term? You need to clean up the brackets, I can think of three different expressions here

$\displaystyle \displaystyle \lim_{x \to 0} \cos(x) - \dfrac{x}{2}$ is direct substitution

$\displaystyle \displaystyle \lim_{x \to 0} \cos(x) - \dfrac{1}{2x}$. Think about what happens when you divide by a number smaller than 1 - the cos(x) term will be negligible in comparison to the 1/(2x) term

$\displaystyle \displaystyle \lim_{x \to 0} \cos \left(x - \dfrac{x}{2}\right) = \lim_{x \to 0} \cos \left(\frac{x}{2}\right)$ which is direct subbing again

4. ## Re: Cosine Trig Limit

Is this $\displaystyle \displaystyle \lim_{x \to 0}\frac{\cos{x} - 1}{2x}$? If so

\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\cos{x} - 1}{2x} &= \frac{1}{2}\lim_{x \to 0}\frac{\cos{x} - 1}{x} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{(\cos{x} - 1)(\cos{x} + 1)}{x(\cos{x} + 1)} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{\cos^2{x} - 1}{x(\cos{x} + 1)} \\ &= \frac{1}{2}\lim_{x \to 0}\frac{-\sin^2{x}}{x(\cos{x} + 1)} \\ &= -\frac{1}{2}\lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0}\frac{\sin{x}}{\cos{x} + 1}\\ &= -\frac{1}{2}\cdot 1 \cdot \frac{0}{2} \\ &= 0 \end{align*}

5. ## Re: Cosine Trig Limit

Normally, I complain about people using L'Hopital's rule when simpler methods will work but I think that is the best way to handle this one:
$\displaystyle \lim_{x\to 0}\frac{cos(x)-1}{2x}= \lim_{x\to 0}\frac{-sin(x)}{2}= 0$

Or use a power series: $\displaystyle cos(1)= 1- (1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot$ so
$\displaystyle \frac{cos(x)-1}{2x}= \frac{-(1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot}{2x}= -(1/4)x+ (1/48)x^3+ \cdot\cdot\cdot$
which goes to 0 as x goes to 0.

6. ## Re: Cosine Trig Limit

Originally Posted by HallsofIvy
Normally, I complain about people using L'Hopital's rule when simpler methods will work but I think that is the best way to handle this one:
$\displaystyle \lim_{x\to 0}\frac{cos(x)-1}{2x}= \lim_{x\to 0}\frac{-sin(x)}{2}= 0$

Or use a power series: $\displaystyle cos(1)= 1- (1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot$ so
$\displaystyle \frac{cos(x)-1}{2x}= \frac{-(1/2)x^2+ (1/24)x^4+ \cdot\cdot\cdot}{2x}= -(1/4)x+ (1/48)x^3+ \cdot\cdot\cdot$
which goes to 0 as x goes to 0.
Or just multiply top and bottom by the top's conjugate, like I did...

7. ## Re: Cosine Trig Limit

Originally Posted by HallsofIvy
Normally, I complain about people using L'Hopital's rule when simpler methods will work but I think that is the best way to handle this one:
$\displaystyle \lim_{x\to 0}\frac{cos(x)-1}{2x}= \lim_{x\to 0}\frac{-sin(x)}{2}= 0$
Alternatively, one could just use the definition of the derivative.

$\displaystyle \lim_{x \to 0} \frac{\cos(x) - 1}{x} = \left. \frac d {dx} \cos(x) \right|_{x = 0} = -\sin(0) = 0.$

I guess this is probably begging the question a little bit, though, since we effectively need to know this limit to calculate the derivatives of sin and cos in the first place. If you already have $\displaystyle \frac{\sin(x)}{x} \to 1$ as $\displaystyle x \to 0$ and you are working with the geometric definitions (not the power series definitions), then Prove It's proof seems best so far.

8. ## Re: Cosine Trig Limit

Originally Posted by purplec16
limx->0 cosx-1/2x

ok i kinda dont know where to start
Please use brackets to make it obvious what the limit is of, as it is everyone helping you may be wasting their time by making the wrong guess at what you are really trying to ask.

CB