Question

Solve ${lim}_{x\to 0}\frac{4^x-9^x}{x(4^x+9^x)}$

• A. $\log \left(\frac{2}{3}\right)$
• B. $\frac{1}{2}\log \left(\frac{3}{2}\right)$
• C. $\frac{1}{2}\log \left(\frac{2}{3}\right)$
• D. $\log \left(\frac{3}{2}\right)$

Answer 1

The correct option is A $\log \left(\frac{2}{3}\right)$

$\underset{x\to 0}{lim}\frac{4^x-9^x}{x(4^x+9^x)}=f\left(x\right)$

By applying $L-$Hospital Rule,

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{\frac{d}{dx}{4}^x-9^x}{\frac{d}{dx}\left(x\right(4^x+9^x\left)\right)}$

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{4^x\ln 4-9^x\ln 9}{4^x+\left(4^x\ln 4\right)x+9^x+x{9}^x\ln 9}$

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{\ln 4-\ln 9}{1+1}$

$\therefore \underset{x\to 0}{lim}\frac{4^x-9^x}{x(4^x+9^x)}=\ln \frac{2}{3}$