Lazy Limits 7!

Express $\begin{array}{rl} y &= \lim_{x \rightarrow 0} \left(\cos(x)\right)^{1/x^2}\\ \ln(y) &= \lim_{x \rightarrow 0} \dfrac{1}{x^2} \ln\left(\cos(x)\right)\\ &= \lim_{x \rightarrow 0} \dfrac{\ln\left(\cos(x)\right)}{x^2} \end{array}$ Since as $x \rightarrow 0$ , $\dfrac{\ln\left(\cos(x)\right)}{x^2} \rightarrow \dfrac{0}{0}$ . In this case, apply L'Hôpital's Rule to get $\begin{array}{rl} \lim_{x \rightarrow 0} \dfrac{\frac{d}{dx}\left[\ln\left(\cos(x)\right)\right]}{\frac{d}{dx}\left[x^2\right]} &= \lim_{x \rightarrow 0} \dfrac{-\frac{\sin(x)}{\cos(x)}}{2x}\\ &= \lim_{x \rightarrow 0} -\dfrac{\sin(x)}{2x\cos(x)}\\ &= -\dfrac{1}{2} \lim_{x \rightarrow 0} \dfrac{\sin(x)}{x\cos(x)} \end{array}$ Because $\lim_{x\rightarrow 0} f(x)g(x) = \lim_{x\rightarrow 0} f(x) \cdot \lim_{x\rightarrow 0}g(x)$ where $f(x) = \dfrac{\sin(x)}{x}$ and $g(x) = \dfrac{1}{\cos(x)}$ , we express the limit as $-\dfrac{1}{2} \left(\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x}\right)\left(\lim_{x \rightarrow 0} \dfrac{1}{\cos(x)}\right)$ Since $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1 \quad \text{and} \quad \lim_{x \rightarrow 0} \dfrac{1}{\cos(x)} = 1$ $\begin{array}{rl} \ln(y) &= -\dfrac{1}{2} \left(\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x}\right)\left(\lim_{x \rightarrow 0} \dfrac{1}{\cos(x)}\right)\\ &= -\dfrac{1}{2} \cdot 1 \cdot 1\\ \ln(y) &= -\dfrac{1}{2}\\ y &= e^{-1/2} = \dfrac{1}{\sqrt{e}} \end{array}$ Thus, $\boxed{e^{-1/2} \approx 0.606}$ .

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0} (\cos x)^{\frac 1{x^2}} & \small \color{#3D99F6}\text{A }1^\infty \text{ case, use } \lim_{x \to a} \left(\frac {f(x)}{g(x)}\right)^{h(x)} = e^{\lim_{x \to a} h(x) \left(\frac {f(x)}{g(x)}-1\right)} \\ & = \exp \left(\lim_{x \to 0} \frac {\cos x -1}{x^2} \right) & \small \color{#3D99F6}\text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{x \to 0} \frac {-\sin x}{2x} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \exp \left(\lim_{x \to 0} \frac {-\cos x}{2} \right) & \small \color{#3D99F6} \text{Differentiate up and down again} \\ & = e^{-\frac 12} = \frac 1{\sqrt e} \approx \boxed{0.607} \end{aligned}$