Question

STATEMENT-1 : $\underset{x\to \alpha }{lim}\frac{\sin \left(f\right(x\left)\right)}{x-\alpha },$ where $f\left(x\right)=a{x}^2+bx+c$, is finite and non-zero, then $\underset{x\to \alpha }{lim}\frac{e^{\frac{1}{f\left(x\right)}}-1}{e^{\frac{1}{f\left(x\right)}}+1}$ does not exist.
STATEMENT-2 : $\underset{x\to \alpha }{lim}\frac{f\left(x\right)}{x-\alpha }$ can take finite value only when it takes $\frac{0}{0}$ form.

• 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

The correct option is A STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

For $\underset{x\to \alpha }{lim}\frac{\left(f\right(x\left)\right)}{x-\alpha }$, the denominator tends to 0; hence, the numerator must also tend to 0 for limit to be finite.

Then, $\alpha$ is a root of the equation $a{x}^2+bx+c=0$ or

$f\left(\alpha \right)=0$.
Also, consider $f\left({\alpha }^{+}\right)\to 0^{+}$ and $f\left({\alpha }^{-}\right)\to 0^{-},$ i.e.,

$\underset{x\to {\alpha }^{+}}{lim}\frac{e^{1/f\left(x\right)}}{e^{1/f\left(x\right)}+1}=\underset{x\to {\alpha }^{+}}{lim}\frac{1-e^{-1/f\left(x\right)}}{1+e^{-1/f\left(x\right)}}=1$
and $\underset{x\to {\alpha }^{-}}{lim}\frac{e^{1/f\left(x\right)}}{e^{1/f\left(x\right)}+1}=1$
Thus, both the statements are true and statement 2 is the correct explanation of statement 1
Hence, option 'A' is correct.