Question

$\underset{x\to 0}{lim}\frac{a^x+a^{-x}-2}{x^2}$ is equal to

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

Answer 1

Answer: (A)

$(\log a{)}^2$

$\underset{x\to 0}{lim}\frac{a^x+a^{-x}-2}{x^2}$

$=\underset{x\to 0}{lim}\frac{a^x\log a-a^{-x}\log a}{2x}........$[Applying L'Hospital's rule]

$=\underset{x\to 0}{lim}\frac{a^x(\log a{)}^2+a^{-x}(\log a{)}^2}{2}$ [Again, applying L'Hospital's rule]

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

$=(\log a{)}^2$