A) If\[\underset{x\to c}{\mathop{\lim }}\,f(x).g(x)\]and\[\underset{x\to c}{\mathop{\lim }}\,f(x),\]then\[\underset{x\to c}{\mathop{\lim }}\,g(x)\]exists
B) If\[\underset{x\to c}{\mathop{\lim }}\,f(x).g(x)\]exists, then\[\underset{x\to c}{\mathop{\lim }}\,f(x)\]and \[\underset{x\to c}{\mathop{\lim }}\,f(x)\underset{x\to c}{\mathop{\lim }}\,g(x)\]exists
C) If\[\underset{x\to c}{\mathop{\lim }}\,\{f(x)+g(x)\}\]and\[\underset{x\to c}{\mathop{\lim }}\,f(x)\]exists, then \[\underset{x\to c}{\mathop{\lim }}\,g(x)\]exists
D) If\[\underset{x\to c}{\mathop{\lim }}\,\{f(x)+g(x)\}\]exists, then\[\underset{x\to c}{\mathop{\lim }}\,f(x)\]and \[\underset{x\to c}{\mathop{\lim }}\,g(x)\]exists
Correct Answer: C
Solution :
(a) This is false, \[f(x)=x;g(x)=\frac{1}{{{e}^{x}}-1}\] Now, \[\underset{x\to 0}{\mathop{\lim }}\,f(x).g(x)\]exist \[=1\] Also, \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=0\]exists but\[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]does not exists. (b) This is false. Let\[f\]be defined as\[f(x)=\left\{ \begin{matrix} 1, & if & x\le 0 \\ 2, & if & x>0 \\ \end{matrix} \right.\]let\[g(x)=0\] Then, \[f(x).g(x)\]and so, \[\underset{x\to 0}{\mathop{\lim }}\,f(x).g(x)\]exists, while\[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]does not. (c) This is true, as\[g=(f+g)-f\]. Therefore, by the limit theorem, \[\underset{x\to 0}{\mathop{\lim }}\,g(x)=\underset{x\to 0}{\mathop{\lim }}\,\{f(x)+g(x)\}-\underset{x\to 0}{\mathop{\lim }}\,f(x)\] (d) This is false.
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