How do I find the limit of x as it approaches zero

[UrbanXrisis]

How do I find the limit of x as it approaches zero of sin(3x)/x? I seriously have no clue what to do here

 

[arildno]

Set 3x=u, so that sin(3x)/x=3*sin(u)/u

 

[UrbanXrisis]

3sin(u)/u is the limit?

 

[cookiemonster]

\lim_{u \to 0} 3 \frac{\sin{u}}{u} = 3\lim_{u \to 0} \frac{\sin{u}}{u} = 3 \cdot 1 = 3

cookiemonster

 

[UrbanXrisis]

how did the three get placed outside the ()?

How do I find the limit of x as it approaches zero of [1-sqrt(x^2+1)]/x^2?

I sub in u so it becomes [1-sqrt(u^2+1)]/u^2...?

 

[arildno]

Okay:
\frac{\sin(3x)}{x}=3(\frac{\sin(3x)}{3x})

We now set u(x)=3x, and consider the function G(x)=F(u(x)), where:
F(u)=3\frac{sin(u)}{u}

We are interested in computing \lim_{x\to{0}}G(x)

For a composite function with continuous kernel u and "shell" F at x=x_{0}, we have:
\lim_{x\to{x}_{0}}F(u(x))=F(\lim_{x\to{x}_{0}}u(x))=\lim_{u\to{u}_{0}}F(u), u_{0}=u(x_{0})

The limit of \frac{\sin{u}}{u} as u goes to zero, is 1, as the Muffinangel states.

 

[e(ho0n3]

The limit of \frac{\sin{u}}{u} as u goes to zero, is 1, as the Muffinangel states.

Wait a moment, how exactly is

\lim_{u \to 0}\frac{\sin u}{u} = 1​

I can see how this is true from a graphical perspective (if I graph sin u and u and look at their behaviour close to zero, or if I draw an appropriate triangle and see how it behaves as one of the angles goes to zero), but is there any ways of determining this algebraically.

 

[cookiemonster]

L'Hospital's rule.

\lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0} \frac{\cos{x}}{1} = 1

Urban's second problem can be solved by L'Hospital's, as well.

cookiemonster

Edit: Who would'a thunk it... There's an s in L'Hospital!

 

[e(ho0n3]

L'Hospital's rule.

\lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0} \frac{\cos{x}}{1} = 1

Urban's second problem can be solved by L'Hospital's, as well.

cookiemonster

Edit: Who would'a thunk it... There's an s in L'Hospital!

That is cheating since you need to now the derivative of sin x to use L'Hopital's rule and to compute the derivative of sin x you need to figure out that limit. You can't use taylor series either since that uses derivates as well. I need a more fundamental explanation.

 

[AKG]

Wait a moment, how exactly is

\lim_{u \to 0}\frac{\sin u}{u} = 1​

I can see how this is true from a graphical perspective (if I graph sin u and u and look at their behaviour close to zero, or if I draw an appropriate triangle and see how it behaves as one of the angles goes to zero), but is there any ways of determining this algebraically.

For one, you could use l'hopital's rule, but I think there should be a way to do it without that.

The proof from my book essentially states that as we approach zero, \theta < \tan \theta and that \frac{\sin \theta}{\theta} < 1, so we get \cos \theta < \frac{\sin \theta}{\theta} < 1 and by the Squeeze Theorem the limit is proved.

 

[cookiemonster]

All you need to know is that the derivative of sin(x) at 0 is 1. You don't need the actual derivative, no?

cookiemonster

 

[arildno]

"Proof" of the limit \lim_{\theta\to0}\frac{\sin\theta}{\theta}

Consider the three regions in the plane:
a) A (unit-) circular section with angle \theta
b) The inscribed right-angle triangle with sides \sin\theta,\cos\theta,1
c) The "outer" triangle with short sides given by 1,\tan\theta

Obviously, the respective areas A, B, C fulfill B<=A<=C
So, we have, using the radian angle measure:
\frac{1}{2}\sin\theta\cos\theta\leq\frac{1}{2}\theta\leq\frac{1}{2}\tan\theta

Or:
\cos\theta\leq\frac{\theta}{\sin\theta}\leq\frac{1}{\cos\theta}

Taking the limit to zero angle yields by the Squeeze theorem, as AKG says, the result.

 

[HallsofIvy]

"How do I find the limit of x as it approaches zero of [1-sqrt(x^2+1)]/x^2?"

Multiply numerator and denominator by 1+ sqrt(x^2+1) and see what happens.