JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer Let\[f(x)=\frac{x-\left\{ x+1 \right\}}{x-\left\{ x+2 \right\}}\]; where\[\{x\}\]is the fractional part of\[x\], then\[\underset{x\to 1/3}{\mathop{\lim }}\,f(x)\]

    A)  has value\[0\]             

    B)  has value\[1\]

    C)  has value\[-\infty \]                 

    D)  has value\[\infty \]

    Correct Answer: B

    Solution :

     As\[x\to \frac{1}{3};\,\,\{x+1\}\to \{1+1/3\}\to 1/3\] Similarly\[\{x+2\}\to \frac{1}{3}\]as\[x\to \frac{1}{3}\]\[\Rightarrow \]\[\underset{x\to 1/3}{\mathop{\lim }}\,f(x)=\underset{x\to 1/3}{\mathop{\lim }}\,\frac{x-1/3}{x-1/3}=1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner