Question

If $\underset{x\to a}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}$ exists, then

• A. both $\underset{x\to a}{lim}f\left(x\right)$ and $\underset{x\to a}{lim}g\left(x\right)$ must exist
• B. $\underset{x\to a}{lim}f\left(x\right)$ need not exist but $\underset{x\to a}{lim}g\left(x\right)$ exists
• C. neither $\underset{x\to a}{lim}f\left(x\right)$ nor $\underset{x\to a}{lim}g\left(x\right)$ may exist
• D. $\underset{x\to a}{lim}f\left(x\right)$ exists but $\underset{x\to a}{lim}g\left(x\right)$ need not exist

Answer 1

Answer: (C)

neither $\underset{x\to a}{lim}f\left(x\right)$ nor $\underset{x\to a}{lim}g\left(x\right)$ may exist

For option A, take $f\left(x\right)=x$ and $g\left(x\right)=\frac{1}{x}$.
Limit of $g\left(x\right)$ doesn't exist at $x=0$, but for $\underset{x\to 0}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}$ limit exists.
For option B, take $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$,
Limits of $g\left(x\right)$ and $f\left(x\right)$ limit doesn't exist as $x\to 0$ individually, but for $\underset{x\to 0}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}$ exists.
For option C, consider the above stated example where $\underset{x\to 0}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}$, but limits of neither $f\left(x\right)$ nor $g\left(x\right)$ may exist.
For option D, consider $f\left(x\right)=x^2$ and $g\left(x\right)=x$, where $\underset{x\to 0}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}$ exists and limit of $f\left(x\right)$ and $g\left(x\right)$ also exists at $x=0$ which contradicts option D.
Hence option C.