A) 2
B) -1
C) 0
D) 1
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\left\{ \frac{1}{x}-\frac{2}{{{e}^{2x}}-1} \right\}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{2x}}-1-2x}{x({{e}^{2x}}-1)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{e}^{2x}}-2}{({{e}^{2x}}-1)+2x{{e}^{2x}}}\] (using L' Hospital rule) \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{4{{e}^{2x}}}{4{{e}^{2x}}+4x{{e}^{2x}}}=1\] \[\therefore \] \[f(x)\]is continuous at\[x=0,\]then \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[1=f(0)\]You need to login to perform this action.
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