Question

Find the limits of $\frac{a^x-b^x}{x},$ when $x=0$.

Answer 1

To find the limits of $\frac{a^x-b^x}{x}$, when $x=0$

i.e. $\underset{x\to 0}{lim}\frac{a^x-b^x}{x}$

$=\underset{x\to 0}{lim}\frac{a^x-1-b^x+1}{x}$

$=\underset{x\to 0}{lim}\frac{(a^x-1)-(b^x-1)}{x}$

$=\underset{x\to 0}{lim}\frac{(a^x-1)}{x}-\underset{x\to 0}{lim}\frac{(b^x-1)}{x}$

Since $\underset{x\to 0}{lim}\frac{(a^x-1)}{x}={\log }_{e}a$

$\therefore \underset{x\to 0}{lim}\frac{(a^x-1)}{x}-\underset{x\to 0}{lim}\frac{(b^x-1)}{x}$

$={\log }_{e}a-{\log }_{e}b$

$={\log }_e\left(\frac{a}{b}\right)$