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I'm sure i'm just missing the trick here, would appreciate some help. I tried lhopital and it didn't help, perhaps a trig identity?

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nofe
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2 Answers2

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Taylor series are the "right" tool to use. Why the right tool? Because the Taylor expansion of $f(x)$ about $x=0$ gives detailed information about the local behaviour of $f(x)$ near $0$.

However, if you want to use L'Hospital's Rule, proceed as follows. The top and bottom clearly approach $0$. Note that $(\cos x)^{\sin x}=\exp(\sin x \ln(\cos x))$.

If the limit of the ratio of the derivatives exists, so does our limit. The derivative of the top is $$\left(\frac{-\sin^2 x}{\cos x} +(\ln(\cos x))(\cos x)\right) \exp(\sin x \ln(\cos x)) .$$ Things are beginning to look messy! However, the factor $\exp(\sin x \ln(\cos x))$ approaches $1$, so it is safe, we can forget about it. Quickly we end up wanting the limit of $$\frac{\frac{-\sin^2 x}{\cos x} +(\ln(\cos x))(\cos x)}{3x^2}.$$ Two mechanical uses of L'Hospitals's Rule now give the answer. We might fiddle a bit with the expression before differentiating, to make life easier. For example, we can multiply top and bottom by $\cos x$, then forget about the $\cos x$ at the bottom.

But we can save sweat in various ways, for example by noting that it is a familiar fact that $(\sin^2 x)/(x^2)$ approaches $1$, so now we need only worry about $(\ln(\cos x))/(3x^2)$.

André Nicolas
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compute the taylor sery of $\cos(x)^{\sin(x)}$ by passing throught exp. You get at $0$ $$\cos(x)^{\sin(x)}=1-\frac{x^3}{2}+\frac{x^6}{8}+o(x^6).$$

Louis La Brocante
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    I agree that the Taylor series approach is the "right" way to approach many of these problems. Using L'Hospital's Rule usually makes a student lose contact with the ground. – André Nicolas Jan 12 '12 at 20:08
  • what does "passing throught exp." mean? how do i calculate the taylor expansion of cos(x)^sin(x) ? – nofe Jan 13 '12 at 13:47