Lista limite indeterminado

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CAI- Limite Indeterminado :Lista de Exercício 1                                                                                                                                                                                                O material que deseja acessar está em código tex, que precisa de alguns segundos para carregar.Em caso de erro nas expressões matemáticas, atualize a página :D Calcule os seguintes Limites Resposta: $ \displaystyle\frac{1}{12}$ $e^{2} $ Ver resposta em vídeo passo a passo clique aqui! 01 $\mathrm{} \displaystyle\lim_{ x \to \infty } \left( 1+\displaystyle\frac{2}{x} \right)^{x} $ $ \displaystyle\frac{1}{e} $ 02 $\mathrm{} \displaystyle\lim_{ x \to \infty } \left( 1- \displaystyle\frac{1}{x} \right)^{x}$ $\displaystyle\frac{1}{e}$ 03 $ \displaystyle\lim_{ x \to \infty } \left( \displaystyle\frac{x}{1+x} \right)^{x} $ $e$ 04 $ \displaystyle\lim_{ \xi \to \infty } \left( 1+\displaystyle\frac{1}{\xi} \right)^{\xi+5}$ $1$ 05 $ \displaystyle\lim_{ \varphi \to \infty } \left\{ \varphi[ \ln{(\varphi + 1)} -\ln{(\varphi)} ] \right\} $ $e^{3}$ 06 $ \displaystyle\lim_{ x \to \displaystyle\frac{\pi}{2} } \left( 1+\cos(x)\right)^{3\mbox{sec}(x)} $ $\alpha$ 07 $ \displaystyle\lim_{ x \to 0 }\displaystyle\frac{ \ln{(1+\alpha x)} }{x} $ $e$ 08 $ \displaystyle\lim_{ x \to \infty } \left( \displaystyle\frac{2x+3}{2x+1} \right)^{x+1} $ $e^{3}$ 09 $\displaystyle\lim_{ x \to 0 } \left( 1+3\mbox{tg}^{2}(x) \right)^{\mbox{cotg}^{2}(x) } $ $1$ para $ \varphi \to +\infty , $ $ 0$ para $ \varphi \to -\infty$ 10 $ \displaystyle\lim_{ \varphi \to \infty } \displaystyle\frac{ \ln{(1+e^{\varphi})}}{\varphi} $ $-\displaystyle\frac{1}{8} $ 11 $ \displaystyle\lim_{ \varphi \to \displaystyle\frac{\pi}{2} } \displaystyle\frac{ \ln{(\mbox{sen}(\varphi))}}{(\pi -2\varphi)^{2} } $ $0$ 12 $\displaystyle\lim_{ \varphi \to \infty } \displaystyle\frac{ \ln{(\varphi)}}{\varphi^{n}} $ ( Em que $n>0$ ) $1$ 13 $ \displaystyle\lim_{ \varphi \to \infty } \displaystyle\frac{ \ln{\left( 1+\displaystyle\frac{1}{\varphi} \right)}}{\mbox{arccotg}(\varphi) } $ $-1$ 14 $\displaystyle\lim_{ \varphi \to \infty } \displaystyle\frac{ \ln{\displaystyle\left(\frac{\varphi+1}{\varphi} \right) }} {\ln{\displaystyle\left(\frac{\varphi-1}{\varphi} \right)}} $ $0$ 15 $\displaystyle\lim_{ \varphi \to 0 } \left[ \varphi- \varphi^{2}\displaystyle\ln{\left(1+\displaystyle\frac{1}{\varphi}\right)} \right] $ $\displaystyle\frac{1}{e}$ 16 $ \displaystyle\lim_{ \varphi \to 1 } \left[ \mbox{tg}\left( \displaystyle\frac{\pi \varphi}{4}\right)^{\mbox{tg}\left( \displaystyle\frac{\pi \varphi}{4}\right)} \right] $ $0$ 17 $\displaystyle\lim_{ \varphi \to 0 } \displaystyle\frac{ \ln^{2}{(1+\varphi)} - \mbox{sen}^{2}(\varphi)} {1-\displaystyle e^{-\varphi^{2}}} $ Resposta: $\displaystyle\frac{1}{2}$ Ver resposta em vídeo passo a passo clique aqui! $\mathrm{(a)}\displaystyle\lim_{\xi \to 2}\displaystyle\frac{4\xi^{3}-2\xi^{2}+\xi}{3\xi^{2}+2\xi}$ $\displaystyle\frac{1}{4}$ $\mathrm{(b)}\displaystyle\lim_{x\to 0} \displaystyle\frac{x}{x(\sqrt{x+4}+2)} $ $0$ $\mathrm{(c)}\displaystyle\lim_{x\to 0} \displaystyle\frac{x^2}{x(\sqrt{x+4}+2)} $ $$ $\mathrm{(d)} \displaystyle\lim_{x\to 0} \displaystyle\frac{1}{x\sqrt{1+x}} -\displaystyle\frac{1}{x} $ $$ $\mathrm{(e)} \displaystyle\lim_{x \to 0} \displaystyle\frac{\sqrt[3]{x+7}-2}{x-1} -\displaystyle\frac{1}{x} $ $$ $\mathrm{(f)} \displaystyle\lim_{x\to 0} \mathrm{arctg}\left(\displaystyle\frac{x^2-4}{3x^2-6x}\right) $ $$ $\mathrm{(g)} \displaystyle\lim_{\varpi \to 2} \displaystyle\frac{\sqrt{\varpi+2}-2}{\sqrt[4]{\varpi+2}-\sqrt{2}}$ $4$ $\mathrm{(h)} \displaystyle\lim_{\psi \to 1} \displaystyle\frac{\sqrt{\psi}-1}{\sqrt[6]{\psi}-1}$ $1 $ $\mathrm{(i)} $ $e$ $\mathrm{(j)} $ $\displaystyle e^{2}$ $\mathrm{(k)} $ $\displaystyle\frac{2}{3}$ $\mathrm{(l)} $ $\displaystyle\frac{2}{3}$ $\mathrm{(m)} $ $\displaystyle 3$ $\mathrm{(n)} $ $2$ $\mathrm{(o)} $ $\omega$ $\mathrm{(p)} $ $-\displaystyle\frac{1}{4}$ $\mathrm{(q)} $                                                     [1] PISKOUNOV. Cálculo Diferencial e Integral, vol I e II. Editora Lopes da Silva.

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