Question

$\displaystyle \lim_{x \to 0} \frac{(\cos x)^{\csc^2 x} - e^{\csc^2 x (\cos x -1)}}{x^2}$

Answer 1

$\displaystyle -\frac{1}{8\sqrt{e}}$

Answer 2

The problem asks to evaluate the limit: $L = \lim_{x \to 0} \frac{(\cos x)^{\csc^2 x} - e^{\csc^2 x (\cos x -1)}}{x^2}$ This limit is of the form $\frac{0}{0}$. We will use Taylor series expansions for $\cos x$ and $\sin x$ around $x=0$. We need expansions up to $x^4$ for the numerator to get a non-zero constant term after dividing by $x^2$.

The Taylor series expansions for $\cos x$ and $\sin x$ are: $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6)$ $\sin x = x - \frac{x^3}{3!} + O(x^5) = x - \frac{x^3}{6} + O(x^5)$ From these, we can find the expansion for $\sin^2 x$: $\sin^2 x = \left(x - \frac{x^3}{6} + O(x^5)\right)^2 = x^2 - 2 \cdot x \cdot \frac{x^3}{6} + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$

Let's evaluate the first term in the numerator, $A = (\cos x)^{\csc^2 x}$. We take the natural logarithm: $\ln A = \csc^2 x \ln(\cos x) = \frac{\ln(\cos x)}{\sin^2 x}$ First, expand $\ln(\cos x)$. Let $u = \cos x - 1$. $u = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - 1 + O(x^6) = -\frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$ Using the expansion $\ln(1+u) = u - \frac{u^2}{2} + O(u^3)$: $\ln(\cos x) = \ln(1+u) = \left(-\frac{x^2}{2} + \frac{x^4}{24}\right) - \frac{1}{2}\left(-\frac{x^2}{2}\right)^2 + O(x^6)$ $\ln(\cos x) = -\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^4}{8} + O(x^6) = -\frac{x^2}{2} + \left(\frac{1}{24} - \frac{3}{24}\right)x^4 + O(x^6)$ $\ln(\cos x) = -\frac{x^2}{2} - \frac{2x^4}{24} + O(x^6) = -\frac{x^2}{2} - \frac{x^4}{12} + O(x^6)$ Now substitute this into the expression for $\ln A$: $\ln A = \frac{-\frac{x^2}{2} - \frac{x^4}{12} + O(x^6)}{x^2 - \frac{x^4}{3} + O(x^6)} = \frac{-\frac{1}{2} - \frac{x^2}{12} + O(x^4)}{1 - \frac{x^2}{3} + O(x^4)}$ Using the approximation $\frac{1}{1-y} = 1+y+y^2+O(y^3)$ for small $y$: $\ln A = \left(-\frac{1}{2} - \frac{x^2}{12}\right)\left(1 + \frac{x^2}{3} + \left(\frac{x^2}{3}\right)^2\right) + O(x^6)$ $\ln A = \left(-\frac{1}{2} - \frac{x^2}{12}\right)\left(1 + \frac{x^2}{3} + \frac{x^4}{9}\right) + O(x^6)$ $\ln A = -\frac{1}{2} - \frac{x^2}{6} - \frac{x^4}{18} - \frac{x^2}{12} - \frac{x^4}{36} + O(x^6)$ $\ln A = -\frac{1}{2} + \left(-\frac{1}{6} - \frac{1}{12}\right)x^2 + \left(-\frac{1}{18} - \frac{1}{36}\right)x^4 + O(x^6)$ $\ln A = -\frac{1}{2} + \left(-\frac{2}{12} - \frac{1}{12}\right)x^2 + \left(-\frac{2}{36} - \frac{1}{36}\right)x^4 + O(x^6)$ $\ln A = -\frac{1}{2} - \frac{3x^2}{12} - \frac{3x^4}{36} + O(x^6)$ $\ln A = -\frac{1}{2} - \frac{x^2}{4} - \frac{x^4}{12} + O(x^6)$ Now, $A = e^{\ln A} = e^{-1/2 - x^2/4 - x^4/12 + O(x^6)} = e^{-1/2} e^{-x^2/4 - x^4/12 + O(x^6)}$. Using the approximation $e^y = 1+y+\frac{y^2}{2!} + O(y^3)$: $A = e^{-1/2} \left(1 + \left(-\frac{x^2}{4} - \frac{x^4}{12}\right) + \frac{1}{2!}\left(-\frac{x^2}{4}\right)^2 + O(x^6)\right)$ $A = e^{-1/2} \left(1 - \frac{x^2}{4} - \frac{x^4}{12} + \frac{x^4}{32} + O(x^6)\right)$ $A = e^{-1/2} \left(1 - \frac{x^2}{4} + \left(-\frac{8}{96} + \frac{3}{96}\right)x^4 + O(x^6)\right)$ $A = e^{-1/2} \left(1 - \frac{x^2}{4} - \frac{5x^4}{96} + O(x^6)\right)$

Now let's evaluate the exponent of the second term in the numerator, $K = \csc^2 x (\cos x -1)$. $K = \frac{\cos x - 1}{\sin^2 x} = \frac{\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^6)}{x^2 - \frac{x^4}{3} + O(x^6)} = \frac{-\frac{1}{2} + \frac{x^2}{24} + O(x^4)}{1 - \frac{x^2}{3} + O(x^4)}$ $K = \left(-\frac{1}{2} + \frac{x^2}{24}\right)\left(1 + \frac{x^2}{3} + \left(\frac{x^2}{3}\right)^2\right) + O(x^6)$ $K = \left(-\frac{1}{2} + \frac{x^2}{24}\right)\left(1 + \frac{x^2}{3} + \frac{x^4}{9}\right) + O(x^6)$ $K = -\frac{1}{2} - \frac{x^2}{6} - \frac{x^4}{18} + \frac{x^2}{24} + \frac{x^4}{72} + O(x^6)$ $K = -\frac{1}{2} + \left(-\frac{1}{6} + \frac{1}{24}\right)x^2 + \left(-\frac{1}{18} + \frac{1}{72}\right)x^4 + O(x^6)$ $K = -\frac{1}{2} + \left(-\frac{4}{24} + \frac{1}{24}\right)x^2 + \left(-\frac{4}{72} + \frac{1}{72}\right)x^4 + O(x^6)$ $K = -\frac{1}{2} - \frac{3x^2}{24} - \frac{3x^4}{72} + O(x^6)$ $K = -\frac{1}{2} - \frac{x^2}{8} - \frac{x^4}{24} + O(x^6)$ Let $B = e^{\csc^2 x (\cos x -1)} = e^K$. $B = e^{-1/2 - x^2/8 - x^4/24 + O(x^6)} = e^{-1/2} e^{-x^2/8 - x^4/24 + O(x^6)}$ $B = e^{-1/2} \left(1 + \left(-\frac{x^2}{8} - \frac{x^4}{24}\right) + \frac{1}{2!}\left(-\frac{x^2}{8}\right)^2 + O(x^6)\right)$ $B = e^{-1/2} \left(1 - \frac{x^2}{8} - \frac{x^4}{24} + \frac{x^4}{128} + O(x^6)\right)$ $B = e^{-1/2} \left(1 - \frac{x^2}{8} + \left(-\frac{1}{24} + \frac{1}{128}\right)x^4 + O(x^6)\right)$ $B = e^{-1/2} \left(1 - \frac{x^2}{8} - \frac{13x^4}{384} + O(x^6)\right)$

Now, substitute $A$ and $B$ back into the limit expression: $L = \lim_{x \to 0} \frac{e^{-1/2} \left(1 - \frac{x^2}{4} - \frac{5x^4}{96}\right) - e^{-1/2} \left(1 - \frac{x^2}{8} - \frac{13x^4}{384}\right) + O(x^6)}{x^2}$ $L = \lim_{x \to 0} \frac{e^{-1/2} \left[\left(1 - \frac{x^2}{4} - \frac{5x^4}{96}\right) - \left(1 - \frac{x^2}{8} - \frac{13x^4}{384}\right)\right] + O(x^6)}{x^2}$ $L = \lim_{x \to 0} \frac{e^{-1/2} \left[-\frac{x^2}{4} + \frac{x^2}{8} - \frac{5x^4}{96} + \frac{13x^4}{384}\right] + O(x^6)}{x^2}$ $L = \lim_{x \to 0} \frac{e^{-1/2} \left[\left(-\frac{2}{8} + \frac{1}{8}\right)x^2 + \left(-\frac{20}{384} + \frac{13}{384}\right)x^4\right] + O(x^6)}{x^2}$ $L = \lim_{x \to 0} \frac{e^{-1/2} \left[-\frac{x^2}{8} - \frac{7x^4}{384}\right] + O(x^6)}{x^2}$ Divide by $x^2$: $L = \lim_{x \to 0} e^{-1/2} \left[-\frac{1}{8} - \frac{7x^2}{384}\right] + O(x^4)$ As $x \to 0$, the terms with $x^2$ and higher powers go to zero. $L = e^{-1/2} \left(-\frac{1}{8}\right) = -\frac{1}{8\sqrt{e}}$

The final answer is $\boxed{-\frac{1}{8\sqrt{e}}}$.