Question

Let $f:R$ $\to$ $R$ be such that $f\left(a\right)=1,f^{\prime }\left(a\right)=2$. If $\underset{x\to 0}{lim}{\left(\frac{f^3(a+x)}{f\left(a\right)}\right)}^{\frac{1}{x}}=e^{\lambda }$, then $\lambda$ is greater than.

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

${lim}_{x\to 0}{\left(\frac{f^3(a+x)}{f\left(a\right)}\right)}^{\frac{1}{x}}=e^{\lambda }$

Taking $\log$ on both sides of the equation

${lim}_{x\to 0}\left(\frac{3\log f(a+x)-\log f\left(a\right)}{x}\right)=\lambda$

Given, $f\left(a\right)=1$

${lim}_{x\to 0}\left(\frac{3\log f(a+x)}{x}\right)=\lambda$ This is Equation (1)

We know that $f(a+x)=x{f}^{\prime }\left(a\right)+f\left(a\right)$

$3{lim}_{x\to 0}\frac{\log [x{f}^{\prime }\left(a\right)+f\left(a\right)]}{x}=\lambda$

Putting the values of

$f^{\prime }\left(a\right)=2$

$f\left(a\right)=1$

$3{lim}_{x\to 0}\frac{\log [2x+1]}{x}=\lambda$ This is Equation (2)

Rewrite ${lim}_{x\to 0}\frac{\log [2x+1]}{x}as{lim}_{x\to 0}\left(\frac{\ln (2x+1)}{x}\right)$

${lim}_{x\to 0}\left(\frac{\ln (2x+1)}{x}\right)$

Apply L'Hospital's Rule

$={lim}_{x\to 0}\left(\frac{\frac{2}{2x+1}}{1}\right)$

Refine $={lim}_{x\to 0}\left(\frac{2}{2x+1}\right)$

Plug in the value $x=0$

$=\frac{2}{2\cdot 0+1}$

Simplify $=2$

From equation (2), we have

$3\times 2=\lambda$

$\lambda =6$