# Homework Help: How do I find the limit of x as it approaches zero

1. Jun 8, 2004

### 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

2. Jun 8, 2004

### arildno

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

3. Jun 8, 2004

### UrbanXrisis

3sin(u)/u is the limit?

4. Jun 8, 2004

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

5. Jun 8, 2004

### 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...?

6. Jun 8, 2004

### 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 continous 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.

7. Jun 8, 2004

### e(ho0n3

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.

8. Jun 8, 2004

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.

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

Last edited: Jun 8, 2004
9. Jun 8, 2004

### e(ho0n3

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.

10. Jun 8, 2004

### AKG

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.

11. Jun 8, 2004

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

12. Jun 9, 2004

### 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.

13. Jun 9, 2004

### 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.