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Posted by **mike** on Monday, December 6, 2010 at 7:59pm.

lim (cos(x))^(7/x^2) as x goes x-->0^+

- L'hospital rule -
**MathMate**, Monday, December 6, 2010 at 8:16pmL'Hôpital's rule applies when there is a fraction whose numerator and denominator are both undefined or zero.

Since the given expression is not a fraction, we need to transform it to a form where L'Hôpital's rule applies.

Taking log is a good way when powers are involved:

ln((cos(x))^(7/x^2))

=(7/x²)*ln(cos(x))

=7ln(cos(x))/x²

Now both numerator and denominator become zero as x->x+.

differentiate with respect to x:

7(-sin(x)/cos(x)) / 2x

=-7tan(x)/2x

Since the expression is still undefined when x->0+, we can differentiate again:

-7sec²(x)/2

=-7/2 as x->0+

So the original limit can be found by raising -7/2 to the power of e, or

Lim x->0+ cos(x)^(7/x²)

= e^{-7/2}

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