Question

If $\underset{x\to \infty }{lim}\left(\ln \right(x^2+5x)-2\ln (cx+1\left)\right)=-2$, then

• A. $c=e^{-1}$
• B. $c=e^{-2}$
• C. $c=e$
• D. $c=e^{-\frac{1}{2}}$

Answer 1

Answer: (C)

$c=e$

Given that, $\underset{x\to \infty }{lim}(\ln (x^2+5x)-2\ln (cx+1))=-2$

$⟹\underset{x\to \infty }{lim}(\ln (x^2+5x)-\ln {(cx+1)}^2)=-2$

$⟹\underset{x\to \infty }{lim}\ln \left(\frac{x^2+5x}{c^2{x}^2+1+2cx}\right)=-2$ ....(since, $\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b}\right)$)

$⟹\underset{x\to \infty }{lim}\ln \left(\frac{x^2(1+5\frac{1}{x})}{x^2(c^2+\frac{1}{x^2}+2c\frac{1}{x^2})}\right)=-2$

$⟹\underset{x\to \infty }{lim}\ln \left(\frac{(1+5\frac{1}{x})}{(c^2+\frac{1}{x^2}+2c\frac{1}{x^2})}\right)=-2$

$⟹\ln \left(\frac{(1+5\left(0\right))}{(c^2+0+2c\left(0\right))}\right)=-2$

$⟹\ln \left(\frac{1}{c^2}\right)=-2$

$⟹\ln \left(1\right)-\ln \left(c^2\right)=-2$

$⟹\ln \left(c^2\right)=2$

$⟹e^2=c^2$

$⟹c=e$