Evaluate the limit and justify your answer.
(a)
\[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} =\answer {e} \]
Justification:
Notice that \(e^{\sin {x}} =f(g(x)) \), where
\[ f(x)=\answer {e^{x}} \hspace {0.2in}and \hspace {0.2in} g(x)=\answer {\sin {x}} \]
and both functions, \(f\) and \(g\), are not continuouscontinuous on \( (-\infty ,\infty )\).
So, we can apply CompositionProduct limit law
\[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} = e^{\lim _{x\to \frac {\pi }{2}}\answer {\sin {x}}} \]
Since \(g\) is not continuouscontinuous at \(\frac {\pi }{2}\), we have
\[ \lim _{x\to \frac {\pi }{2}}\sin {x}=\sin {\Bigl (\answer {\frac {\pi }{2}}\Bigr )}=\answer {1} \]
and, therefore
\[ \lim _{x\to \frac {\pi }{2}}e^{\sin {x}} =e^{\answer {1}} =\answer {e} \]
_________
(b)
\[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} = \answer {-\frac {\sqrt {2}}{2}} \]
Justification:
Notice that \(\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )}=f(g(x)) \), where
\[ f(x)=\answer {\cos {x}} \hspace {0.2in}and \hspace {0.2in} g(x)=\answer {\pi \frac {3x^{2}-4x+1}{4}} \]
and both functions, \(f\) and \(g\), are not continuouscontinuous on \( (-\infty ,\infty )\).
So, we can apply CompositionProduct limit law
\[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} =\cos {\Bigl (\lim _{x\to 2}\answer {\pi \frac {3x^{2}-4x+1}{4}}\Bigr )} \]

Since \(g\) is not continuouscontinuous at \(2\), we have

\[ \lim _{x\to 2}\pi \frac {3x^{2}-4x+1}{4}=\pi \frac {3\answer {2}^{2}-4\answer {2}+1}{4}=\answer {\frac {5\pi }{4}} \]
and, therefore
\[ \lim _{x\to 2}\cos {\Bigl (\pi \frac {3x^{2}-4x+1}{4}\Bigr )} =\cos {\Bigl (\answer {\frac {5\pi }{4}} \Bigr )}= \answer {-\frac {\sqrt {2}}{2}} \]

________________________________________________________________________________________________

(c)
\[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} = \answer {64} \]
Justification: Notice that \((\ln {x}+5)^{2} =f(g(x)) \), where
\[ f(x)=\answer {x^{2}} \hspace {0.2in}and \hspace {0.2in} g(x)=\answer {\ln {x}+5} \]
The function \(f\) is not continuouscontinuous on \( (-\infty ,\infty )\).
The function \(g\) is a productsum of two continuous functions, \(\ln {x}\) and \( \answer {5}\). Therefore, \(g\) is not continuouscontinuous on its domain \((\answer {0},\answer {\infty })\).
So, we can apply CompositionProduct limit law
\[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} =\Bigl (\lim _{x\to e^{3}}(\ln {x}+5)\Bigr )^{\answer {2}} \]
Since \(g\) is not continuouscontinuous at \(e^{3}\), we have
\[ \lim _{x\to e^{3}}(\ln {x}+5)=\ln {(\answer {e^{3}})}+5=\answer {3}+5=\answer {8} \]
and, therefore
\[ \lim _{x\to e^{3}}(\ln {x}+5)^{2} = \Bigl (\answer {8}\Bigr )^{2} =\answer {64} \]
_________