ok so this is what it says to do

evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0

2. Hello, runner07!

Remember this theorem? . . $\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

Evaluate: .$\displaystyle \lim_{x\to0^+}\frac{\sin x}{x^{\frac{1}{2}}}$

We have: .$\displaystyle \frac{\sin x}{\sqrt{x}}$

Multiply by $\displaystyle \frac{\sqrt{x}}{\sqrt{x}}\!:\;\;\frac{\sqrt{x}}{\s qrt{x}}\cdot\frac{\sin x}{\sqrt{x}} \;=\;\sqrt{x}\cdot\frac{\sin x}{x}$

Then: .$\displaystyle \lim_{x\to0^+}\left[\sqrt{x}\cdot\frac{\sin x}{x}\right] \;=\;\sqrt{0}\cdot1 \;=\;0$

3. OH! i forgot about that theorem. but that makes more sense. thank you!!

4. ## wat

evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0[/quote]

5. Originally Posted by kritica

evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0
When using L'Hopital's rule, you will always be dividing by zero since the power of one half will never disappear. Follow the eloquent solution that Soroban provided.

EDIT: Actually, never mind, I had an algebraic brain fart! Look at topsquark's post below!

6. Originally Posted by kritica

evaluate lim (sinx)/ (x^(1/2)) as x goes to 0+

ok so i've used LHopital's rule but what i got was dne. i think i did something wrong but i dont know what... here's what i did

lim (cosx)/((1/2) x ^(-1/2)) as x--> 0+ = 1/0
It can be done by L'Hopital's rule. But watch that negative exponent in the denominator.

$\displaystyle \lim_{x \to 0^+} \frac{sin(x)}{\sqrt{x}}$

$\displaystyle = \lim_{x \to 0^+} \frac{cos(x)}{\frac{1}{2\sqrt{x}}}$
ala L'Hopital

$\displaystyle = \lim_{x \to 0^+} 2 \sqrt{x} ~ cos(x) = 0$

-Dan

7. Also it is known for very small values of x $\displaystyle \sin(x)\sim{x}$

which means you can replace the $\displaystyle \sin(x)$ with x in this case givign

$\displaystyle \lim_{x\to{0}}\frac{x}{\sqrt{x}}=\lim_{x\to{0}}\sq rt{x}=0$