Answer to Question #223356 in Calculus for cayyy

Question

Find limit using L' Hopitals rule:

(a) lim x $\to$ 0 (cos2x)^(1/x^2)

Answer 1

(a) $lim_{x\rightarrow 0} (cos2x)^{\frac{1}{x^2}}$

$\because u=e^{ln(u)}$

$\implies lim_{x\rightarrow0}(cos2x)^{1/x^2}=lim_{x\rightarrow0}e^{ln(cos2x)^{1/x^2}}=lim_{x\rightarrow0}e^{\frac{ln(cos2x)}{x^2}}$

$\implies$ Since we have an intermediate of the type $\dfrac{0}{0}$ , So we can use L'Hopital's Rule -

$\implies e^{lim_{x\rightarrow0}\frac{cos2x}{x^2}}=e^{lim_{x\rightarrow0}\frac{\frac{d}{dx}(ln(cos2x))}{\frac{d}{dx}(x^2)}} = e^{lim_{x\rightarrow0}\frac{(-sin2x)}{xcos2x}} = e^{lim_{x\rightarrow0}(\frac{-tan2x}{x})}$

$\implies e^{-lim_{x\rightarrow0}(\frac{tan2x}{x})}$

Again, Since we have an intermediate of the type $\dfrac{0}{0}$ , So again we can use l'Hopital's Rule -

$\implies e^{-lim_{x\rightarrow0}(\frac{tan2x}{x})}$ $= e^{-lim_{x\rightarrow 0}(\frac{d/dx(tan2x)}{d/dx(x)})}=e^{-lim_{x\rightarrow 0}(2sec^22x)}$

$\\\implies$ $e^{-lim_{x\rightarrow 0}(2sec^22x)} = e^{-2\times 1}=e^{-2}$

Hence, $lim_{x\rightarrow 0 }(cos2x)^{\frac{1}{x^2}}=\dfrac{1}{e^2}$