Find the following limit \[ \lim _{x \rightarrow 0^{+}}(4 x)^{x} \] What must be done to put the given expression in a form to which l'Hôpital's rule can be directly applied?

Step 1 ：Given the limit expression \(\lim _{x \rightarrow 0^{+}}(4 x)^{x}\), we notice that as x approaches 0, the expression takes the form of an indeterminate form 0^0.

Step 2 ：To apply l'Hôpital's rule, we need to transform the expression into the form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Step 3 ：We can achieve this by taking the natural logarithm of the expression, which allows us to bring the exponent x in front. Let's denote y = (4x)^x and take the natural logarithm of both sides. We get \(\ln y = x \ln(4x)\).

Step 4 ：Now, we can find the limit of \(\ln y\) as x approaches 0 from the right, and then exponentiate the result to find the limit of y.

Step 5 ：The limit of \(\ln y\) as x approaches 0 from the right is -\(\infty\). Therefore, the limit of y is \(e^{-\infty} = 0\).

Step 6 ：Final Answer: The limit of \((4x)^x\) as x approaches 0 from the right is \(\boxed{0}\).