- \[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} =\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}} \] Justificativa:
Observe que \(e^{\sin {x}} =f(g(x)) \), em que \[ f(x)=\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}} \hspace {0.2in}and \hspace {0.2in} g(x)=\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}} \] e ambas as funções, \(f\) e \(g\), são não contínuascontínuas em \( (-\infty ,\infty )\).
Portanto, podemos aplicar a lei da composiçãodo produto para limites \[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} = e^{\lim _{x\to \frac {\pi }{2}}\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}}} \] Como \(g\) é não contínuacontínua em \(\frac {\pi }{2}\), temos \[ \lim _{x\to \frac {\pi }{2}}\sin {x}=\sin {\Bigl (\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}}\Bigr )}=\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}} \] e, portanto, \[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} =e^{\class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}}} =\class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}} \] ________________________________________________________________________________________________ -
\[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} = \class{xm-answer}{\cssId{xm-answer-9}{\phantom{\rule{7em}{1.5em}}}} \] Justificativa:
Observe que \(\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )}=f(g(x)) \), em que \[ f(x)=\class{xm-answer}{\cssId{xm-answer-10}{\phantom{\rule{7em}{1.5em}}}} \hspace {0.2in}and \hspace {0.2in} g(x)=\class{xm-answer}{\cssId{xm-answer-11}{\phantom{\rule{7em}{1.5em}}}} \] e ambas as funções, \(f\) e \(g\), são não contínuascontínuas em \( (-\infty ,\infty )\).
Portanto, podemos aplicar a lei da composiçãodo produto para limites \[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} =\cos {\Bigl (\lim _{x\to 2}\class{xm-answer}{\cssId{xm-answer-12}{\phantom{\rule{7em}{1.5em}}}}\Bigr )} \]Como \(g\) é não contínuacontínua em \(2\), temos \[ \lim _{x\to 2}\pi \frac {3x^{2}-4x+1}{4}=\pi \frac {3\class{xm-answer}{\cssId{xm-answer-13}{\phantom{\rule{7em}{1.5em}}}}^{2}-4\class{xm-answer}{\cssId{xm-answer-14}{\phantom{\rule{7em}{1.5em}}}}+1}{4}=\class{xm-answer}{\cssId{xm-answer-15}{\phantom{\rule{7em}{1.5em}}}} \] e, portanto, \[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} =\cos {\Bigl (\class{xm-answer}{\cssId{xm-answer-16}{\phantom{\rule{7em}{1.5em}}}} \Bigr )}= \class{xm-answer}{\cssId{xm-answer-17}{\phantom{\rule{7em}{1.5em}}}} \]
_____________________________________________________________
- \[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} = \class{xm-answer}{\cssId{xm-answer-18}{\phantom{\rule{7em}{1.5em}}}} \] Justificativa: Observe que \((\ln {x}+5)^{2} =f(g(x)) \), em que \[ f(x)=\class{xm-answer}{\cssId{xm-answer-19}{\phantom{\rule{7em}{1.5em}}}} \hspace {0.2in}and \hspace {0.2in} g(x)=\class{xm-answer}{\cssId{xm-answer-20}{\phantom{\rule{7em}{1.5em}}}} \] A função \(f\) é não contínuacontínua
em \( (-\infty ,\infty )\).
A função \(g\) é um produtouma soma de duas funções contínuas, \(\ln {x}\) e \( \class{xm-answer}{\cssId{xm-answer-21}{\phantom{\rule{7em}{1.5em}}}}\). Portanto, \(g\) é não contínuacontínua em seu domínio \((\class{xm-answer}{\cssId{xm-answer-22}{\phantom{\rule{7em}{1.5em}}}},\class{xm-answer}{\cssId{xm-answer-23}{\phantom{\rule{7em}{1.5em}}}})\).
Logo, podemos aplicar a lei da composiçãodo produto para limites \[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} =\Bigl (\lim _{x\to e^{3}}(\ln {x}+5)\Bigr )^{\class{xm-answer}{\cssId{xm-answer-24}{\phantom{\rule{7em}{1.5em}}}}} \] Como \(g\) é não contínuacontínua em \(e^{3}\), temos \[ \lim _{x\to e^{3}}(\ln {x}+5)=\ln {(\class{xm-answer}{\cssId{xm-answer-25}{\phantom{\rule{7em}{1.5em}}}})}+5=\class{xm-answer}{\cssId{xm-answer-26}{\phantom{\rule{7em}{1.5em}}}}+5=\class{xm-answer}{\cssId{xm-answer-27}{\phantom{\rule{7em}{1.5em}}}} \] e, portanto, \[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} = \Bigl (\class{xm-answer}{\cssId{xm-answer-28}{\phantom{\rule{7em}{1.5em}}}}\Bigr )^{2} =\class{xm-answer}{\cssId{xm-answer-29}{\phantom{\rule{7em}{1.5em}}}} \] ________________________________________________________________________________________________
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