USING L'HOPITAL'S RULE, find the limit (CLICK TO SEE)

Question

amenah8 · Answer

limit as x-->0+ of (x^2)/sinx-s

misty1212 · Answer

HI!!

amenah8 · Answer

sinx-X, sorry!

amenah8 · Answer

hello!

misty1212 · Answer

$$\lim_{x\to 0^+}\frac{x^2}{\sin(x)}$$?

misty1212 · Answer

looks like you put a $-s$ in the denominator, is that there?

amenah8 · Answer

except on the denominator: sinx-x

misty1212 · Answer

oooh

misty1212 · Answer

so you probably have to use l'hopital twice since you will get $\frac{0}{0}$ the first time

amenah8 · Answer

yeah, but then i got 2/0 and that still doesn't work

misty1212 · Answer

first time gives $$\frac{2x}{\cos(x)-1}$$ right ?

amenah8 · Answer

yeah, that was 0/0, then i used the rule again and got 2/0

misty1212 · Answer

ok fine, that means the two sided limit does not exist
either goes to $\infty$ or $-\infty$ depending

amenah8 · Answer

so if it is approaching zero from the positive side, i plug in 0.000001 for x?

misty1212 · Answer

that is probably why they wrote $x\to 0^+$ so you can figure out which

misty1212 · Answer

you don't have to do that,but it will work, maybe easier just to think

misty1212 · Answer

you get $$\frac{2}{-\sin(x)}$$ the second time right?

misty1212 · Answer

if $x>0$ but close to $0$ then $\sin(x)>0$ so $-\frac{2}{\sin(x)}<0$

amenah8 · Answer

right

michele_laino · Answer

hint:
another procedure, is to expand the function at denominator with Taylor's expantion:
$$\Large  \frac{{{x^2}}}{{\sin x - x}} = \frac{{{x^2}}}{{ - \frac{{{x^3}}}{6} + o\left( {{x^3}} \right)}} = \frac{{ - 6}}{{x + o\left( x \right)}}$$

misty1212 · Answer

therefore it goes to $-\infty$ not $\infty$

michele_laino · Answer

expansion*

amenah8 · Answer

oh! i understand. thank you so much!!!

michele_laino · Answer

:)