Question

Statement $I:\underset{x\to 0}{\text{lim}}\left[x\right]\left\{\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right\}$ (where [.] represents the greatest integer function) does not exist

Statement $II:\underset{x\to 0}{\text{lim}}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)$ does not exist

• A. Both $I$ and $II$ are individually true and $II$ is the correct explanation of $I$
• B. Both $I$ and $II$ are individually true but $II$ is not the correct explanation of $I$
• C. $I$ is true but $II$ is false
• D. $I$ is false but $II$ is true

Answer 1

The correct option is B Both $I$ and $II$ are individually true but $II$ is not the correct explanation of $I$

$\underset{x\to 0^{+}}{\text{lim}}\left[x\right]\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)=\underset{h\to 0}{\text{lim}}\left[h\right]\left(\frac{1-e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}\right)=0\times 1=0$

$\underset{x\to 0^{-}}{\text{lim}}\left[x\right]\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)=\underset{h\to 0}{\text{lim}}[-h]\left(\frac{e^{\frac{1}{-h}}-1}{e^{\frac{1}{-h}}+1}\right)=-1\times (-1)=1$

Thus, given limit does not exist.
Also, $\underset{x\to 0}{\text{lim}}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)$ does not exist, but this cannot be reason for non-existence of $\underset{x\to 0}{\text{lim}}\left[x\right]\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right).$