1. ## limits

i have to find the limit of this

lim x->0+ $sin(2x)/x^2$

once i simplify this i get $2/x$

and i get stuck. can someone help? thanks

2. Originally Posted by break
i have to find the limit of this

lim x->0+ $sin(2x)/x^2$

once i simplify this i get $2/x$

and i get stuck. can someone help? thanks
Note the indeterminant form $\frac{0}{0}$. This implies that L'Hopital's rule applies.

3. so lim x->0+ $sin(2x)/x^2$

then using L'Hopital's rule,

lim x->0+ $cos(2x)/2x$ = 1

is this correct?

4. Originally Posted by break
so lim x->0+ $sin(2x)/x^2$

then using L'Hopital's rule,

lim x->0+ $cos(2x)/2x$ = 1

is this correct?
Not Quite

$\lim_{x\to0^+}\frac{\sin{2x}}{x^2}=\lim_{x\to0^+}\ frac{2\cos{2x}}{2x}=\lim_{x\to0^+}\frac{\cos{2x}}{ x}=\infty$

5. why does $cos(2x)/x$ = infinity?

can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?

6. Originally Posted by break
why does $cos(2x)/x$ = infinity?

can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?
How would multiplying 2 to cos(2X) cancel it out?
Also note, Von Nemo corrected your incorrect differentiation when you applied L'Hoptials Rule

7. Originally Posted by RockHard
How would multiplying 2 to cos(2X) cancel it out?
Also note, Von Nemo corrected your incorrect differentiation when you applied L'Hoptials Rule
$cos(2x)/2x * 2/1$

cause

$cos(2x)/2x = 1$ am i right? :/

8. Originally Posted by break
why does $cos(2x)/x$ = infinity?

can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?
You're thinkinging of (sinx)/x. This limit $\to1$ as $x\to0$. So, your thought would be spot on if we were talking about sine, but we're concerned with cosine here.

9. Originally Posted by VonNemo19
You're thinkinging of (sinx)/x. This limit $\to1$ as $x\to0$. So, your thought would be spot on if we were talking about sine, but we're concerned with cosine here.
so that rule only applies to sin(x)? no other trig functions?

10. Originally Posted by break
so that rule only applies to sin(x)? no other trig functions?
Yes. $\lim_{x\to0}\frac{\sin{x}}{x}=1$ is a special limit derived for the purposes of determining the trigonometric derivatives.

Check out the proof here...

World Web Math: Useful Trig Limits

11. Originally Posted by break
i have to find the limit of this

lim x->0+ $sin(2x)/x^2$

once i simplify this i get $2/x$

and i get stuck. can someone help? thanks
Substitute the first few terms of the Maclaurin series for sin(2x) and the limit becomes obvious.

12. Hello, i'm also studying limits right now!

I hope you get why you must use l'hopital. (It's 0/0)
Cos 2x is 2 when it reaches 0+ and 2x reaches 0, and 2/0 is infinity , that's why the solution is infinity.

13. kev91: $cos2x \rightarrow 1$ as $x \to 0^+$, not 2, but the answer remains the same.

Another way to do this without L'hopital is $sin(2x) = 2sinxcosx \Rightarrow \frac{sin(2x)}{x^2} = 2 \frac{sinx}{x} \cdot \frac{cosx}{x}$, and now since $\frac{cosx}{x} \to \infty \text{ as } x \to 0^+$, we get $\lim_{x \to 0^+} 2 \cdot \frac{sinx}{x} \cdot \frac{cosx}{x} = 2 \cdot \lim_{x \to 0^+} \frac{cosx}{x} = \infty$

14. When you have 0/infinity can you say equals zero? And if you have infinity/0 can you say equals infinity? Or do you have to use l'hopital?

15. Informally speaking, yes -- when you have a fraction with numerator reaching zero and denominator reaching infinity, the fraction reaches zero. The other statement is true as well : really big number divided by really small number is like really big number times another really big number, equals really really big number!

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