Real Limit Problem Found in a Scholarship Exam for a Masters C0urse

Question

Evaluate: $\lim_{x\rightarrow0}(\cos x)^{\frac{1}{\sin^2x}}$.

I found this problem in one of the previous years papers of a Scholarship exam I'll be giving soon. Any hints are welcome!

Answer 1

Hint:

Introduce substitution $$\sin x=\frac1{u}$$

You will get a limit which looks like the famous definition of $e$.

The result is $\frac1{\sqrt e}$. This is all fairly simple, it's better to try yourself and learn something along the way.

Answer 2

When $1>\cos x >0$ we have $\ln ((\cos x)^{(1/\sin^2 x)})=$ $\frac {1}{\sin^2 x}\ln \cos x=$ $\frac {1}{u^2}\ln \sqrt {1-u^2}$ where $u=\sin x.$

Answer 3

$$\cos(x)^{\frac{1}{\sin^2(x)}} = \exp\left(\frac{\ln\cos(x)}{\sin^2(x)}\right) = \exp\left(\frac{\ln\cos(x)}{\tfrac12[1-\cos(2x)]}\right)$$

Dealing with the fractional expression, recall the approximations of $\cos$ and $\ln$ centered around $0$ and $1$, respectively: $$\cos(x) = 1 - x^2/2 +O(x^4)$$

$$\ln (1-x) = -x+O(x^2)$$ then for the numerator we get \begin{align} \ln\cos(x) &= \ln(1-x^2/2+O(x^4))\\ &= -(x^2/2-O(x^4)) + O\big((x^2/2-O(x^4))^2\big)\\ &= -\tfrac12x^2 + O(x^4) \end{align} Hence, by continuity of exponentiation, \begin{align} \lim_{x\to 0} \cos(x)^{\frac{1}{\sin^2(x)}} &= \exp\left( \lim_{x\to 0} \frac{2\ln\cos(x)}{1-\cos(2x)} \right)\\ &=\exp\left( \lim_{x\to 0} \frac{2\big(-\tfrac12 x^2 + O(x^4)\big)}{1 - \big(1-(2x)^2/2+O((2x)^4)\big)} \right)\\ &=\exp\left( \lim_{x\to 0} \frac{-x^2 + O(x^4)}{2x^2+O(x^4)} \right) =\exp\left( -\tfrac12 \right)\\ \end{align}