zero divide by zero
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i chanced upon this question, lim (sin x)/x, as x->0.
the solution used geometry i think, but i didnt really understand it, and i think it uses 0/0. this got me thinking to whether 0/0 = 0(where you consider 0 divide by any number = 0) or 1 (where u consider 2 of the same numbers divided together to be 1), or no definite solution (as proven by expressing the division step as an inversion of multiplication step, eg, let 0/0 = a, therefore 0 = a x 0, and in this way, a can be any number.).
besides that, can anyone explain to me why lim (sin x)/x, as x->0 = 1 using the geometry method(it involves the triangle and an arc with the hypoteneuse being the radius of the arc)? (:
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nothing can be divided by 0. r u sure the question is /0?
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Use L'Hopital's rule lah lol....
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This is not a question. It is a standard formula ie lim sinx/x = 1, x > 0 in many maths textbooks.
To see how this formula is derived or proven, type lim sin x/x and the url links will show that lim sinx/x = 1, x > 0 using L' hospital rule.
This formula is usually taught in year 1 poly maths. Students only need to know how to use the formula, proof of the formula is not required.
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Originally posted by jayh272416:
Use L'Hopital's rule lah lol....
Agree with this.
Reminds me of MA1505 in NUS.
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Originally posted by eagle:
Agree with this.
Reminds me of MA1505 in NUS.
MA1505R
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last time me no R
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R stands for what?
Refined?Redo?
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Yeah, I have to agree that L'Hopital's Rule is the only way around this....
0/0 is simply an invalid mathematical operation.
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lol.. 0/0
There is... another way...
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You can also use the Maclaurin expansion.
sin x = x − x3/3! + x5/5! − x7/7! + ...
sin x / x = 1 − x2/3! + x4/5! − x6/7! + ...
→ 1 as x → 0
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this is the geometric proof that i saw, but i have no clue what it means ):
i dont see how when theta->0, sin (theta)/ (theta) -> 1 lol.
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The diagram shows a sector of a circle of radius 1. The angle at the center is θ, in radians. So the arc length in orange is also θ. Cos arc length = radius × angle at center.
The small vertical height on the right is sin θ. Because sin θ = vertical height / hypotenuse. But the hypotenuse is just the radius, 1.
As θ → 0, the arc length approaches the vertical height (more or less obvious).
In other words, θ → sin θ.
In other words, sin θ / θ → 1.
I think that's the meaning of the diagram.
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All these make my head go big.
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oh, so ure supposed to infer form there that theta -> sin theta as theta -> 0 huh.
i kept seeing it as as when theta -> 0, theta -> 0, sin theta -> 0, and the limit of sin(theta)/(theta) -> 0, hence i was wondering if 0/0 = 1...lol.
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Ya deifnitely it's not 0/0 = 1. You cannot divide by 0 in maths, it's not allowed. The reason is the one you suggested above:
no definite solution (as proven by expressing the division step as an inversion of multiplication step, eg, let 0/0 = a, therefore 0 = a x 0, and in this way, a can be any number.)
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absol has a great question that, and I often trip over the notion of limits as well. I think the standard calculus way of "plug and chug" teaching does not help understanding. So just throwing up names like L'Hospital's rule does not help. I believe that learning the "why" is more important than the "how". Calculus is not a set of rules and procedures to follow, it is a collection of beautiful geometric ideas from geniuses like Archimedes and culminating in Newton and Lebniz's deep connection between calculating area under the curve and the tangent problem.
Here is the an article that I found; it takes an intuitive approach to explaining limits
http://betterexplained.com/articles/why-do-we-need-limits-and-infinitesimals/