Limits Continuity And Differentiability Question 41
Question: The function $ f(x) $ defined by $ f(x)= \begin{matrix} {\log _{(4x-3)}}(x^{2}-2x+5),\frac{3}{4}<x<1andx>1 \\ 4,x=1 \\ \end{matrix} . $
Options:
A) is continuous at x=1
B) is discontinuous at x=1 since $ f({1^{+}}) $ does not exist though $ f({1^{-}}) $ exists
C) is discontinuous at x=1 since $ f({1^{-}}) $ does not exist thought $ f({1^{+}}) $ exists
D) is discontinuous at x=1 since neither $ f({1^{+}}) $ nor $ f({1^{-}}) $ exists.
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \underset{x\to {1^{-}}}{\mathop{\lim }}f(x)=\underset{h\to 0}{\mathop{\lim }}f(1-h) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\log (4+h^{2})}{\log (1-4h)}=-\infty $
And $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=\underset{x\to 0}{\mathop{\lim }}f(1+h)=\underset{h\to 0}{\mathop{\lim }}\frac{\log (4+h^{2})}{\log (1+4h)}=\infty $ So, $ f({1^{-}}) $
and $ f({1^{+}}) $ do not exist.
BETA
