1. ## help on limit

1.) lim as theta-->pie/4 tanthetasin^2theta

2.) lim x-->5 from the left 3x-15/|x-5|

thanks

1.) lim as theta-->pie/4 tanthetasin^2theta

2.) lim x-->5 from the left 3x-15/|x-5|

thanks
A. Use brackets to make you question clear, otherwise we are just guessing
what you question realy is.

B. I will assume you mean for your first question:

lim as theta-->pi/4 tan[theta (sin(theta))^2)]

Or:

$\displaystyle \lim_{\theta \to \pi /4} \tan [\theta ( sin( \theta))^2)]$

Now as $\displaystyle \tan [\theta ( sin( \theta))^2)]$ is continuouse at $\displaystyle \pi/4$ the limit may be evaluated by just plugging $\displaystyle \pi/4$ into this to get:

$\displaystyle \lim_{\theta \to \pi /4} \tan [\theta ( sin( \theta))^2)]=\tan [(\pi/4) ( sin( \pi/4))^2)] = \tan (\pi/8)$

C. By " lim x-->5 from the left 3x-15/|x-5|" I will assume you mean:

lim x-->5 from the left (3x-15)/|x-5|

or:

$\displaystyle \lim_{x \to 5 -} \frac{3x-15}{|x-5|}$

$\displaystyle \lim_{x \to 5 -} \frac{3x-15}{|x-5|} = 3 \frac{x-5}{|x-5|}$

but when $\displaystyle x<5$, $\displaystyle \frac{x-5}{|x-5|}=-1$, so:

$\displaystyle \lim_{x \to 5 -} \frac{3x-15}{|x-5|} = 3 \frac{x-5}{|x-5|} = -3$

RonL

The mathematical constant is $\displaystyle \pi$ = "pi", not a dessert.

If I read the first one correctly, it's ridiculous simple . . .

$\displaystyle 1)\;\;\lim_{\theta\to\frac{\pi}{4}}\left[\tan\theta\!\cdot\sin^2\theta\right] \;=\;\tan\left(\frac{\pi}{4}\right)\!\cdot\sin^2\l eft(\frac{\pi}{4}\right) \;=\;1\cdot\left(\frac{1}{\sqrt{2}}\right)^2 \;=\;1\cdot\frac{1}{2} \;=\;\frac{1}{2}$

4. ## hi

thank you all. this forum is so helpful