Question

Evaluate:$\underset{x\to 0}{\lim }\frac{[a^x+a^{-x}-2]}{x^2}$

• A. ${\left(\log a\right)}^2$
• B. $\log a$
• C. $0$
• D. None of these

Answer 1

The correct option is A

${\left(\log a\right)}^2$

Explanation for the correct option:

Finding the value of the given limit:

Simplifying the equation to determinate form and applying the limits:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{[a^x+a^{-x}-2]}{x^2}=\frac{[a^0+a^{-0}-2]}{0^2} \\ =\frac{[a^0+a^{-0}-2]}{0} \\ =\frac{[a^0+a^{-0}-2]}{0} \\ =\frac{[1+1-2]}{0} \\ =\frac{0}{0}\left[\text{Indeterminate Form}\right] \end{gathered}$$

Using the L' hospitality rule

$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{[a^x+a^{-x}-2]}{x^2}=\underset{x\to 0}{\lim }\frac{a^x\ln \left(a\right)-a^{-x}\ln \left(a\right)}{2x}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{a}}^{\mathbf{x}}\mathbf{=}{\mathbf{a}}^{\mathbf{x}}\mathbf{l}\mathbf{n}\left(a\right)\mathbf{]} \\ =\frac{a^0\ln \left(a\right)-a^0\ln \left(a\right)}{2\times 0} \\ =\frac{\ln \left(a\right)-\ln \left(a\right)}{0} \\ =\frac{0}{0}\left[\text{Indeterminate Form}\right] \end{gathered}$$

Again, using the L' hospitality rule

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\ln \left(a\right)\left(a^x-a^{-x}\right)}{2x}=\frac{\ln \left(a\right)}{2}\underset{x\to 0}{\lim }\frac{a^x\ln \left(a\right)-\left(-a^{-x}\ln \left(a\right)\right)}{1}\left[\because \frac{d}{dx}{a}^x=a^x\ln \left(a\right)\right] \\ =\frac{\ln \left(a\right)}{2}\underset{x\to 0}{\lim }\frac{a^x\left(\ln a\right)+a^{-x}\left(\ln a\right)}{1} \\ =\frac{\ln \left(a\right)}{2}\left(\frac{a^0\left(\ln a\right)+a^0\left(\ln a\right)}{1}\right) \\ =\frac{\ln \left(a\right)}{2}\left(\frac{2\times \left(\ln a\right)}{1}\right) \\ =\ln \left(a\right)\times \ln \left(a\right) \\ ={\left(\ln \left(a\right)\right)}^2 \end{gathered}$$

Thus, $\underset{x\to 0}{\lim }\frac{[a^x+a^{-x}-2]}{x^2}={\left(\log a\right)}^2$

Hence, option (A) is the correct answer.