Something simple that i cannot visualize?

[r6mikey]

Something simple that i cannot visualize??

Homework Statement

limit as x approaches 0 for sinx/4x

Homework Equations

ok...what i know is that sinx/x=1 so by that simple statement the answer becomes 1/4 because sinx/x=1 and than there's 4x which gives me the 4...so 1/4...i tend to over think things...another question is why would cos3x=1 when x goes to 0? i know cosx goes to 1 when x equals 0...but wouldn't you multiply 1 by 3 so it would be 3 or is it because it oscillates to 1 all the time?

The Attempt at a Solution

ok so there's a way our teacher does it that I'm trying to figure out that way, even though i know the answer right off the bat, just thinking it might help me in the future

lim as X approaches 0 for sinx/4x...lim as x approaches 0 for (sinx)/((4x/4)*4)...this is to get rid of the 4(or should i be multiplying by 4x?) in the 4x and to put it on the outside...1/4 lim as x approaches 0 for (sinx)/(x)...lim as x approaches 0 for (1/4)*(1)=1/4...look right to everyone? am i cancelling out the 4x right?

 

[Mindscrape]

Hah, you and this other guy must be using the same book or be in the same class. Take a look at the squeeze principle.

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/squeezedirectory/SqueezePrinciple.html

 

[bomba923]

Hint: ignore squeeze theorem, use l'Hôpital's rule

 

[CompuChip]

By L'Hopital,
$$\lim_{x \to 0} \frac{\sin x}{4 x} = \lim_{x \to 0} \frac{\cos x}{4} = \frac{1}{4}$$

Or by Taylor expansion,
$$\lim_{x \to 0} \frac{\sin x}{4 x} = \lim_{x \to 0} \frac{x + \mathcal O(x^3) }{4 x} = \lim_{x \to 0} \frac{1 + \mathcal O(x^2) }{4} = \lim_{x \to 0} \left( \frac{1}{4} + \mathcal O(x^2) \right) = \frac14.$$

Both ways completely analogous to how you would prove the $\sin x/x$ limit.

Or the stupid way
$$\lim_{x \to 0} \frac14 \frac{\sin x}{x} = \left( \lim_{x \to 0} \frac14 \right) \times \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac14 \times 1 = \frac14.$$

 

[bomba923]

Or the stupid way
$$\lim_{x \to 0} \frac14 \frac{\sin x}{x} = \left( \lim_{x \to 0} \frac14 \right) \times \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac14 \times 1 = \frac14.$$