Question

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

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

Answer 1

The correct option is B

$12$

Explanation for the correct option:

Finding the value of the given limit:

$\underset{x\to 0}{\lim }{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}}=l$ (say) ($1^{\infty }$form)

Now,

$$\begin{gathered} l=\underset{x\to 0}{\lim }{\left(\frac{{16}^x+9^x}{2}\right)}^{\frac{1}{x}} \\ =\underset{x\to 0}{\lim }{e}^{\frac{1}{x}\left(\frac{{16}^x+9^x}{2}-1\right)} \\ =exp\underset{x\to 0}{\lim }\left[\frac{1}{x}\left(\frac{{16}^x+9^x}{2}-1\right)\right] \\ =exp\underset{x\to 0}{\lim }\left[\frac{1}{2}\left(\frac{{16}^x+9^x-1-1}{x}\right)\right] \\ =exp\underset{x\to 0}{\lim }\left[\frac{1}{2}\left(\frac{{16}^x-1}{x}+\frac{9^x-1}{x}\right)\right] \\ =exp\underset{x\to 0}{\lim }\left[\frac{1}{2}\left(\ln 16+\ln 9\right)\right]\mathbf{}\mathbf{[}\mathbf{\because }\underset{\mathbf{}\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{a}}^{\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{lna}\mathbf{]} \end{gathered}$$

Applying the limits,

$$\begin{gathered} l=e^{\frac{1}{2}\left(\ln 16+\ln 9\right)} \\ =e^{\frac{1}{2}\ln 16+\frac{1}{2}\ln 9} \\ =e^{\ln {\left(16\right)}^{\frac{1}{2}}+\ln {\left(9\right)}^{\frac{1}{2}}}\left[\because a\ln \left(b\right)=\ln {\left(b\right)}^a\right] \\ =e^{\ln 4+\ln 3} \\ =e^{\ln 12}\left[\because \ln \left(a\right)+\ln \left(b\right)=\ln \left(\mathrm{ab}\right)\right] \\ =12 \end{gathered}$$

Therefore, the value of the given equation is $12$

Hence, option (B) is the correct answer.