## Calculus (3rd Edition)

$$1.8$$
Given $$\lim _{x \rightarrow 0} \frac{12^{x}-1}{4^{x}-1}$$ Consider $$f(x)= \frac{12^{x}-1}{4^{x}-1}$$ From the following figure, we can observe that \begin{align*} \lim _{\theta \rightarrow 0^+} f( x)&=1.8\\ \lim _{\theta \rightarrow 0^-} f( x)&=1.8 \end{align*} Then $$\lim _{x \rightarrow 0} \frac{12^{x}-1}{4^{x}-1}=1.8$$