For the given function \(f\), evaluate the limit and justify your answer.
- (a)
- \(f(x)=x\) \[ \lim _{x\to 7}f(x) = \answer {7} \]Justification:
\(f\) is continuous at \(a=7\), which implies that \(\lim _{x\to 7}f(x)=f(\answer {7})=\answer {7}\)._____________________________________________________________
- (b)
- \(f(x)=x^3\) \[ \lim _{x\to -2}f(x) = \answer {-8} \]Justification:
\(f\) is continuous at \(a=-2\), which implies that
\(\lim _{x\to -2}f(x)=f\Bigl (\answer { -2} \Bigr )=\Bigl (\answer {-2 }\Bigr )^3=\answer {-8}\)._____________________________________________________________
- (c)
- \(f(x)=e^{x}\) \[ \lim _{x\to 0}f(x) = \answer {1} \]Justification:
\(f\) is continuous at \(a=0\), which implies that
\(\lim _{x\to 0}f(x)=f\Bigl (\answer {0}\Bigr )=e^{\answer {0}}=\answer {1}\)._____________________________________________________________
- (d)
- \(f(x)=\ln {x}\) \[ \lim _{x\to e^{4}}f(x) = \answer {4} \]Justification:
\(f\) is continuous at \(a=e^{4}\), which implies that
\(\lim _{x\to e^{4}}f(x)=f\Bigl (\answer {e^{4}}\Bigr )=\ln {\Bigl (\answer {e^{4}}\Bigr )}=\answer {4}\)._____________________________________________________________
- (e)
- \(f(x)=\cos {x}\) \[ \lim _{x\to \frac {2\pi }{3}}f(x) = \answer {-\frac {1}{2}} \]Justification:
\(f\) is continuous at \(a= \frac {2\pi }{3}\), which implies that
\(\lim _{x\to \frac {2\pi }{3}}f(x)=f\Bigl (\answer { \frac {2\pi }{3}}\Bigr )=\cos {\Bigl (\answer { \frac {2\pi }{3}}\Bigr )}=\answer {-\frac {1}{2}}\).