• Nov 28th 2007, 12:07 PM
runner07
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
• Nov 28th 2007, 01:22 PM
Soroban
Hello, runner07!

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

Quote:

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

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

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

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

• Nov 30th 2007, 08:42 PM
runner07
OH! i forgot about that theorem. but that makes more sense. thank you!!
• May 22nd 2008, 11:06 AM
kritica
wat
wat is the answer???????

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]
• May 22nd 2008, 11:10 AM
colby2152
Quote:

Originally Posted by kritica
wat is the answer???????

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!
• May 22nd 2008, 12:23 PM
topsquark
Quote:

Originally Posted by kritica
wat is the answer???????

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.

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

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

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

-Dan
• May 22nd 2008, 12:35 PM
Mathstud28
Also it is known for very small values of x $\sin(x)\sim{x}$

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

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