A) 0
B) 1
C) ?1
D) Does not exist
Correct Answer: D
Solution :
\[f(x)=\left( \frac{{{e}^{1/x}}-1}{{{e}^{1/x}}+1} \right)\,,\] then \[\underset{x\to \,0+}{\mathop{\lim }}\,\,f(x)=\underset{h\to \,0}{\mathop{\lim }}\,\left( \frac{{{e}^{1/h}}-1}{{{e}^{1/h}}+1} \right)=\underset{h\to \,0}{\mathop{\lim }}\,\frac{{{e}^{1/h}}\left( 1-\frac{1}{{{e}^{1/h}}} \right)}{{{e}^{1/h}}\left( 1+\frac{1}{{{e}^{1/h}}} \right)}=1\] Similarly \[\underset{x\to \,0-}{\mathop{\lim }}\,f(x)=-1\]. Hence limit does not exist.
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