question on limit
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pka
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pka
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Well your simplified form can be written as:
\(\displaystyle \dfrac{5\sin(5x)}{5x}\cdot\dfrac{3x}{3\sin(3x)} \cdot \dfrac{1}{\cos(5x)}\)
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There is a theorem that says
\(\displaystyle \displaystyle \lim_{u \rightarrow 0}\dfrac{sin(u)}{u} = 1.\)
That is not a general theorem like L'Hospital's Rule, but highly specific.
It happens, however, to be very useful for this problem.
For a proof, see http://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x
\(\displaystyle \displaystyle \lim_{u \rightarrow 0}\dfrac{sin(u)}{u} = 1.\)
That is not a general theorem like L'Hospital's Rule, but highly specific.
It happens, however, to be very useful for this problem.
For a proof, see http://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x
pka
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Look at reply #4.
\(\displaystyle \displaystyle{\lim _{x \to 0}}\left( {\dfrac{{5\sin (5x)}}{{5x}}} \right) = 5{\lim _{x \to 0}}\left( {\dfrac{{\sin (5x)}}{{5x}}} \right) = 5\)
\(\displaystyle \displaystyle{\lim _{x \to 0}}\left( {\frac{{3x}}{{3\sin (3x)}}} \right) = \frac{1}{3}{\lim _{x \to 0}}\left( {\frac{{3x}}{{\sin (3x)}}} \right) = \frac{1}{3}\)