Question

Assertion :$\underset{x\to 0}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}$ does not exist. Reason: $\underset{x\to 0^{+}}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}$ does not exist.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

R.H.L. $\underset{x\to 0^{+}}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\underset{h\to 0}{lim}\frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}$

$=\underset{h\to 0}{lim}\frac{1-e^{\frac{1}{h}}}{1+e^{\frac{1}{h}}}=\frac{1-e^{-\infty }}{1+e^{-\infty }}=\frac{1-0}{1-0}=1$

$\therefore$ Reason(R) is false.

Again L.H.L. $=\underset{x\to 0^{-}}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\underset{h\to 0}{lim}\frac{e^{-\frac{1}{h}}-1}{e^{-\frac{1}{h}}+1}=\frac{e^{-\infty }-1}{e^{-\infty }+1}=-1$

As L.H.L.$\ne$R.H.L.

$\therefore$ Limit does not exist at $x=0$

$\therefore$ Assertion (A) is true.