Question

STATEMENT-1 : $\underset{x\to 0}{lim}\left[x\right]\left\{\frac{e^{1/x}-1}{e^{1/x}+1}\right\}$ (where [.] represents the greatest integer function) does not exist.
STATEMENT-2 : $\underset{x\to 0}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$ does not exists.

• A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

Answer: (B)

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

I) $\underset{x\to 0}{lim}\left[x\right]\left\{\frac{e^{1/x}-1}{e^{1/x}+1}\right\}$

$RHL=\underset{x\to 0^{+}}{lim}\left[x\right]\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$
$={lim}_{h\to 0}\left[h\right]\left(\frac{e^{1/h}-1}{e^{1/h}+1}\right)$
$={lim}_{h\to 0}\left[h\right]\left(\frac{1-e^{-1/h}}{1+e^{-1/h}}\right)$
$=0\times 1=0$
$LHL=\underset{x\to 0^{-}}{lim}\left[x\right]\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$
$={lim}_{h\to 0}[-h]\left(\frac{e^{-1/h}-1}{e^{-1/h}+1}\right)$
$=-1\times (-1)=1$
Here, $LHL\ne RHL$
Thus, given limit does not exist.
II) $\underset{x\to 0}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$

$RHL=\underset{x\to 0^{+}}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$
$={lim}_{h\to 0}\left(\frac{e^{1/h}-1}{e^{1/h}+1}\right)$
$={lim}_{h\to 0}\left(\frac{1-e^{-1/h}}{1+e^{-1/h}}\right)$
$RHL=1$

$LHL=\underset{x\to 0^{-}}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$
$={lim}_{h\to 0}\left(\frac{e^{-1/h}-1}{e^{-1/h}+1}\right)$
$LHL=-1$

So, $\underset{x\to 0}{lim}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$ does not exist, but this cannot be taken as only reason for non-existence of $\underset{x\to 0}{lim}\left[x\right]\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$.