Thread: lhospital (finals my brain is mush)

1. lhospital (finals my brain is mush)

Why can't I see this I don't know...

lim( as x approaches pi/2 from the left) (secx - tanx)

can't use math tags worth a damn.

Im pretty sure this is lhospital's rule because just taking the limit gives me 1 + infinite which is not the correct answer, I converted with the natural log and did common denominator but I cant think anymore.. studying for 5 finals (calc and physics being 2) just makes my brain melt

2. $\displaystyle \lim_{x \to \frac{\pi}{2}^+} \left( \sec x - \tan x\right) \ = \ \lim_{x \to \frac{\pi}{2}^+}\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x}\right) \ = \ \lim_{x \to \frac{\pi}{2}^+} \frac{1 - \sin x}{\cos x} = \left[\frac{0}{0}\right]$

Now use L'Hopital's.

3. Originally Posted by ryukolink
Why can't I see this I don't know...

lim( as x approaches pi/2 from the left) (secx - tanx)

can't use math tags worth a damn.

Im pretty sure this is lhospital's rule because just taking the limit gives me 1 + infinite which is not the correct answer, I converted with the natural log and did common denominator but I cant think anymore.. studying for 5 finals (calc and physics being 2) just makes my brain melt
$\displaystyle lim_{x \rightarrow \frac{\pi}{2}^-}(\sec x - \tan x)$$\displaystyle =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{1}{\cos x}-\frac{\sin x}{\cos x})$$\displaystyle =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{1-\sin x}{\cos x})$$\displaystyle =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{-\cos x}{-\sin x})$ (Since you got zero over zero)
$\displaystyle =lim_{x \rightarrow \frac{\pi}{2}^-}(\cot x)=?$