# Something simple that i cannot visualize?

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?

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## Answers and Replies

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

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CompuChip
Homework Helper
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
$$\lim_{x \to 0} \frac14 \frac{\sin x}{x} = \left( \lim_{x \to 0} \frac14 \right) \times \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac14 \times 1 = \frac14.$$

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

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