1. ## l' hospital rule

limit (x-pi) (tan (x/2))
x-pi

I found the answer is 1/0.. i think its wrong...
tq..

2. Originally Posted by maiya

limit (x-pi) (tan (x/2))
x-pi

I found the answer is 1/0.. i think its wrong...
tq..
$\lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right]$.

This $\to \frac{0}{0}$, so you can use L'Hospital's Rule.

$\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]$

$= \frac{1}{-\frac{1}{2}}$

$= -2$.

3. Originally Posted by Prove It
$\lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right]$.

This $\to \frac{0}{0}$, so you can use L'Hospital's Rule.

$\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]$

$= \frac{1}{-\frac{1}{2}}$

$= -2$.
tqvm..

4. Originally Posted by Prove It
$\lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right]$.

This $\to \frac{0}{0}$, so you can use L'Hospital's Rule.

$\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]$

$= \frac{1}{-\frac{1}{2}}$

$= -2$.
sorry mr. prove it... can you show me the full way how to get the line 3? i can't get it..

5. Originally Posted by maiya
sorry mr. prove it... can you show me the full way how to get the line 3? i can't get it..
Try substituting $x = \pi$...

6. Do you mean
$
\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]
$
?

ProveIt thought you mean the third equation.

L'Hopital's rule says that, if f(a)= g(a)= 0, then $\lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}$.

In other words, he differentiated the numerator, $(x- \pi)sin\left(\frac{x}{2}\right)$ and the denominator, $cos\left(\frac{x}{2}\right)$, separately.

7. Originally Posted by HallsofIvy
Do you mean
$
\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]
$
?

ProveIt thought you mean the third equation.

L'Hopital's rule says that, if f(a)= g(a)= 0, then $\lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}$.

In other words, he differentiated the numerator, $(x- \pi)sin\left(\frac{x}{2}\right)$ and the denominator, $cos\left(\frac{x}{2}\right)$, separately.
tqvm mr.prove it and HallsofIvy. I got it...