Finding $\lim_{x\to0} \frac{(4^x-1)^3}{\sin\left(\frac{x}{a}\right)\log\left(1+\frac{x^2}{3}\right)}$

Use Taylor series.

As $x \rightarrow 0$:

$$4^x = 1 + x \ln 4 + o(x)$$

$$\sin x = x + o(x)$$ $$\ln(1 + x) = x + o(x)$$

So:

$$\lim_{x\rightarrow 0} \dfrac{(4^x-1)^3}{\sin\left(\dfrac{x}{a}\right)\ln\left(1+\dfrac{x^2}{3}\right)} = \lim_{x\rightarrow 0}\dfrac{(x \ln4)^3}{\left(\frac{x}{a}\right)\left(\frac{x^2}{3}\right)} = 3 a (\ln 4)^3$$

And now time for without Taylor or lhopital as anticipated.

$$L=\lim_{x\rightarrow 0} \dfrac{3 \times a\times \dfrac{(4^x-1)^3}{x^3}}{\dfrac{\sin\left(\dfrac{x}{a}\right)}{\dfrac{x}{a}}\times \dfrac{\log\left(1+\dfrac{x^2}{3}\right)}{\dfrac{x^2}{3}}}$$

Notice how i have written the expression here and it is same as the expression you have mentioned.

We will now use three standard limits given by.

1)$\lim_\limits{x\to 0}\frac{a^x-1}{x}=\ln a$

2)$\lim_\limits{u\to 0}\frac{\sin u}{u}=1$

3)$\lim_\limits{v\to 0}\frac{\ln (1+v)}{v}=1$

Using these limits we can get.

$L=3a \times (\ln 4)^3$