Question

$\underset{x\to 0}{lim}{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}$ is equal to :

• A. $\frac{25}{2}$
• B. $12$
• C. $1$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $12$

To find : $\underset{x\to 0}{lim}{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}$

Applying exponent rule : $a^x=e^{\ln a^x}=e^{x\ln a}$

$\therefore {\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}=e^{\frac{1}{x}\ln \left(\frac{{16}^x+9^x}{2}\right)}$

$=e^{\frac{\ln \left(\frac{{16}^x+9^x}{2}\right)}{x}}$

$\therefore \underset{x\to 0}{lim}{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}=e^{\underset{x\to 0}{lim}\left(\frac{\ln \left(\frac{{16}^x+9^x}{2}\right)}{x}\right)}$

Consider, $L=\underset{x\to 0}{lim}\frac{\left(\frac{{16}^x+9^x}{2}\right)}{x}$

$=\underset{x\to 0}{lim}\frac{\left(\frac{2^{4x}+3^{2x}}{2}\right)}{x}$ which is in the form of $\frac{0}{0}$

So, applying L'Hospital's Rule

$\frac{d}{dx}\left(\frac{2^{4x}+3^{2x}}{2}\right)=\frac{1}{2}\left(2^{4x}\right)\cdot \ln \left(2\right)\cdot 4+3^{2x}\ln \left(3\right)\cdot 2=2^{4x+1}\cdot \ln \left(2\right)+3^{2x}\cdot \ln \left(3\right)$

And $\frac{d}{dx}\left(x\right)=1$

$\therefore L=\underset{x\to 0}{lim}\frac{\left(2^{4x+1}\cdot \ln \left(2\right)+3^{2x}\cdot \ln \left(3\right)\right)}{1}=2\ln \left(2\right)+\ln \left(3\right)$

$\therefore \underset{x\to 0}{lim}{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}=e^{2\ln \left(2\right)+\ln \left(3\right)}=e^{\ln (2{)}^2+\ln (3)}=e^{\ln (2^2\times 3)}=2^2\times 3=12$