Limits Continuity And Differentiability Question 98
Question: Let $ f(x)=g(x).\frac{{e^{1/x}}-{e^{-1/x}}}{{e^{1/x}}+{e^{-1/x}}} $ , where g is a continuous function then $ \underset{x\to 0}{\mathop{\lim }}f(x) $ does not exist if
Options:
A) g(x) is any constant function
B) g(x)=x
C) $ g(x)=x^{2} $
D) g(x) = x h (x), where h(x) is a polynomial.
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to {0^{+}}}{\mathop{\lim }}\frac{{e^{1/x}}-{e^{-1/x}}}{{e^{1/x}}+{e^{-1/x}}}=\underset{x\to {0^{+}}}{\mathop{\lim }}\frac{1-{e^{-2/x}}}{1+{e^{-2}}^{/x}}=1 $ and $ \underset{x\to {0^{-}}}{\mathop{\lim }}\frac{{e^{1/x}}-{e^{-1/x}}}{{e^{1/x}}+{e^{-1/x}}}=\underset{x\to {0^{-}}}{\mathop{\lim }}\frac{{e^{2/x}}-1}{{e^{2/x}}+1}=-1 $
Hence $ \underset{x\to 0}{\mathop{\lim }}f(x) $ exists if $ \underset{x\to 0}{\mathop{\lim }}g(x)=0 $ If $ g(x)=a\ne 0 $ (constant) then $ \underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=a $ and $ \underset{x-{0 _{-}}}{\mathop{\lim }}f(x)=-a $
Thus $ \underset{x\to 0}{\mathop{\lim }}f(x) $ doesn’t exist in this case.
BETA
