Functions Question 6
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{{e^{1/x}}-1}{{e^{1/x}}+1}= $
Options:
A) 0
B) 1
C) -1
D) Does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=( \frac{{e^{1/x}}-1}{{e^{1/x}}+1} ), $ then $ \underset{x\to 0+}{\mathop{\lim }}f(x)=\underset{h\to 0}{\mathop{\lim }}( \frac{{e^{1/h}}-1}{{e^{1/h}}+1} )=\underset{h\to 0}{\mathop{\lim }}\frac{{e^{1/h}}( 1-\frac{1}{{e^{1/h}}} )}{{e^{1/h}}( 1+\frac{1}{{e^{1/h}}} )}=1 $ Similarly $ \underset{x\to 0-}{\mathop{\lim }}f(x)=-1 $ . Hence limit does not exist.
BETA
