Functions Question 505
Question: For the function $ f(x)= \begin{cases} & \frac{{e^{1/x}}-1}{{e^{1/x}}+1},x\ne 0 \\ & 0,x=0 \\ \end{cases} . $ , which of the following is correct
[MP PET 2004]
Options:
A) $ \underset{x\to 0}{\mathop{\lim }},f(x) $ does not exist
B) $ f(x) $ is continuous at $ x=0 $
C) $ \underset{x\to 0}{\mathop{\lim }},f(x)=1 $
D) $ \underset{x\to 0}{\mathop{\lim }},f(x) $ exists but $ f(x) $ is not continuous at $ x=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Given $ f(x)= \begin{cases} & \frac{{e^{\frac{1}{x}}}-1}{{e^{\frac{1}{x}}}+1},x\ne 0 \\ & 0,,,x=0 \\ \end{cases} . $
Þ $ \underset{x\to {0^{+}}}{\mathop{\lim }},\frac{{e^{\frac{1}{x}}}-1}{{e^{\frac{1}{x}}}+1}=\frac{{e^{\infty }}-1}{{e^{\infty }}+1}=-1 $
Þ $ \underset{x\to {0^{-}}}{\mathop{\lim }},\frac{{e^{\frac{1}{x}}}-1}{{e^{\frac{1}{x}}}+1}=\frac{1-{e^{-\frac{1}{x}}}}{1+{e^{\frac{1}{x}}}}=\frac{1-{e^{-\infty }}}{1+{e^{\infty }}}=1 $ So, $ \underset{x\to 0}{\mathop{\lim }},f(x) $ exists at $ x=0 $ , but at $ x=0 $ it is not continuous.
BETA
