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angstrem1

  • 2 years ago

Why does (cos(x)-1)/x go to 0 and sin(x)/x go to 1 as x go to 0? I can see the plots of that functions at wolframalpha, but I can't understand, why do they go to that values? Is there any algebraic proof?

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  1. nickfrei
    • 2 years ago
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    If i understand your question correctly, you're asking why the limit as x tends to zero of (cos(x)-1)/x is zero and sin(x)/x is one. If thats the case then it's because if you plug in numbers for x that are incredibly close to zero (say .001, .0001, .00001 and -.001, -.0001, -.00001) you will find that the closer the x gets to zero the closer the function gets to zero and one for (cos(x)-1)/x and sin(x)/x respectively.

  2. angstrem1
    • 2 years ago
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    nickfrei, thanks for the answer, but I'd like to see if there exists some algebraic proof, the one not involving this 'brute force'.

  3. nickfrei
    • 2 years ago
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    I understand what you're saying and i dont think there is a true algebraic way to solve it. The only thing that comes to mind would be the epsilon delta definition of limit and it's proof which i think may help you out but im not entirely sure about that. I'll provide some links if you want to look up more simply google epsilon delta definition of limit or epsilon delta proof of limit. http://www.math.utah.edu/~petersen/1210/LimitProofs.pdf http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx http://amath.colorado.edu/courses/1340/2009fall/Misc/delta-epsilon.pdf

  4. calculusfunctions
    • 2 years ago
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    We know that\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\]We use this fact to now evaluate\[\lim_{x \rightarrow 0}\frac{ \cos x -1 }{ x }\]Since we know the Pythagorean theorem sin²x + cos²x = 1 or 1 - cos²x = sin²x. To achieve this result, we first multiply both the numerator and the denominator by cos x + 1. Thus\[\lim_{x \rightarrow 0}\frac{ \cos x -1 }{ x }\]\[=\lim_{x \rightarrow 0}(\frac{ \cos x -1 }{ x })(\frac{ \cos x +1 }{ \cos x +1 })\] \[=\lim_{x \rightarrow 0}\frac{ \cos ^{2}x -1 }{ x(\cos x +1) }\] \[=\lim_{x \rightarrow 0}\frac{-(1-\cos ^{2}x) }{ x(\cos x +1) }\] \[=-\lim_{x \rightarrow 0}\frac{ \sin ^{2}x }{ x(\cos x +1) }\] \[=-\lim_{x \rightarrow 0}\frac{ \sin x }{ x }⋅\lim_{x \rightarrow 0}\frac{ \sin x }{ \cos x +1 }\] \[=(-1)(1)(0)\]Therefore\[\lim_{x \rightarrow 0}\frac{ \cos x -1 }{ x }=0\]

  5. ortollj
    • 2 years ago
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    Hello linear approximation f(x)=f(x0)+f'(x0)*(x-x0) http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-approximation-and-curve-sketching/session-23-linear-approximation/

  6. angstrem1
    • 2 years ago
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    Thanks for your answers, I'm sorry for responding so late. calculusfunctions, why does \[\lim_{x \rightarrow 0}\frac{ \sin x }{ \cos x + 1 } = (1)(0)\] and how do we proof, that \[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\] nickfrei, thanks for the proofs of limits, but unfortunately I haven't taken the mathematical analysis course yet, so I still fill a bit uncomfortable with it's theorems (actually, I've used a wrong textbook to learn it and failed...).

  7. calculusfunctions
    • 2 years ago
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    @angstrem1 what I said was\[(-1)∙\lim_{x \rightarrow 0}\frac{ \sin x }{ x }∙\lim_{x \rightarrow 0}\frac{ \sin x }{ \cos x +1 }=(-1)(1)(0)\] To prove that\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\]Let's set up a table of values to evaluate (sin x)/x near 0. Hence in our table, x is any real number near zero, then sin x implies the sine of any angle whose radian measure is x. We will now calculate the on-sided limits.In other words, we will calculate the limit values from the left and the right of zero. as x → 0 from the left x < 0 (sin x)/x 1 0.84147 0.5 0.95885 0.1 0.99833 0.01 0.99998 0.001 0.99999 We see that as x → 0 from the left, (sin x)/x → 1. Thus\[\lim_{x \rightarrow 0^{-}}\frac{ \sin x }{ x }=1\] as x →0 from the right x > 0 (sin x)/x -1 0.84147 -0.5 0.95885 -0.1 0.99833 -0.01 0.99998 -0.001 0.99999 We see that as x → 0 from the right, (sin x)/x → 1. Thus\[\lim_{x \rightarrow 0^{+}}\frac{ \sin x }{ x }=1\]Since\[\lim_{x \rightarrow 0^{-}}\frac{ \sin x }{ x }=1=\lim_{x \rightarrow 0^{+}}\frac{ \sin x }{ x }\]Therefore\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\]

  8. ortollj
    • 2 years ago
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    linear approximation f(x)=f(x0)+f'(x0)*(x-x0) x0= 0 =>sin(x)=sin(0)+cos(0)*x=x then sin(x)/x=1

  9. Stacey
    • 2 years ago
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    |dw:1352270024401:dw| To make the inequality work in all quadrants, then we need the absolute value of each expression.

  10. Stacey
    • 2 years ago
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    |dw:1352271033274:dw| We want as x approaches zero, so we can look at \[-\frac{ \pi }{ 2 } \le x \le \frac{ \pi }{ 2 }\] cos x is positive there and so is (sin x)/x.

  11. Stacey
    • 2 years ago
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    |dw:1352271697457:dw|

  12. mzirino
    • 2 years ago
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    There is a really excellent lecture on this exact subject here: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-8-limits-of-sine-and-cosine/

  13. Brmathmajor
    • 2 years ago
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    I'm not sure which class you are in and what you have learned thus far but both of these are indeterminate forms meaning that the top and bottom approach 0 as x->0. From here we can simply use L'Hopital's rule and take the derivative of the top and bottom. Then you can simply plug in to get the limit.

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