$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$

Solution

$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$

As the fraction is in the form of $$of\ the\ \frac{0}{0}\ form$$

We can use L' Hospitals rule

Differentiating Nr and Dr.

$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$ $$\frac{d}{dx}\left(4^{3x}\ -\ 1\right)\ =\ \left(\ln4.\ 4^{3x}.3\right)\ -\ 0$$

Now again putting limits $$\left(\ln4.\ 4^{3x}.3\right)\ \ =\ \frac{\left(\ln\ 4.4^0.3\right)}{1}\ =\ 3\ln4$$ Answer