lim as x approaches 0 of (sinx/x)^(1/x^2)

Do I take the ln of both sides? then what? :confused:

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- Jun 1st 2006, 04:02 PMSusie38How do I find this limit?
lim as x approaches 0 of (sinx/x)^(1/x^2)

Do I take the ln of both sides? then what? :confused: - Jun 2nd 2006, 12:32 AMTD!
I don't see two sides, but getting a logarithm in there would be handy :)

$\displaystyle

\mathop {\lim }\limits_{x \to 0} \left( {\left( {\frac{{\sin x}}{x}} \right)^{x^{ - 2} } } \right) = \mathop {\lim }\limits_{x \to 0} \left( {\exp \left( {\log \left( {\left( {\frac{{\sin x}}{x}} \right)^{x^{ - 2} } } \right)} \right)} \right)

$

Now we can use a propertie of log to get the exponent up front, the limit can be brought into the exponential. This gives:

$\displaystyle

\mathop {\lim }\limits_{x \to 0} \left( {\exp \left( {\log \left( {\left( {\frac{{\sin x}}{x}} \right)^{x^{ - 2} } } \right)} \right)} \right) = \exp \left( {\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\log \left( {\frac{{\sin x}}{x}} \right)}}{{x^2 }}} \right)} \right)

$

Now you have the case 0/0, so you could apply L'Hopital's rule. One time won't be enough though, and it's not really a fun thing to do. Perhaps easier and certainly as elegant is noticing that we're tending to 0. We could replace the expression with the logarithm with its second order Taylor expansion about 0. The Taylor expansion is:

$\displaystyle

\log \left( {\frac{{\sin x}}{x}} \right) = - \frac{{x^2 }}{6} - \frac{{x^4 }}{{180}} - \cdots

$

Now as said, we can take the second order term only, since the fourth can be neglected arround 0. This gives:

$\displaystyle

\exp \left( {\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \frac{{x^2 }}{6}}}{{x^2 }}} \right)} \right) = \exp \left( {\mathop {\lim }\limits_{x \to 0} \left( { - \frac{1}{6}} \right)} \right) = \exp \left( { - \frac{1}{6}} \right) = e^{ - \frac{1}{6}}

$