Question

find the following limit
lim (5x)^(3x)
x→0+
lim (5x)^(3x) = (type an exact answer.)

Explanation:

Step1: Rewrite using exponential - log relationship

Let $y=(5x)^{3x}$. Then $\ln y = 3x\ln(5x)$. We want to find $\lim_{x
ightarrow0^{+}}\ln y=\lim_{x
ightarrow0^{+}}3x\ln(5x)$. This is in the $0\times\infty$ form. We rewrite it as $\lim_{x
ightarrow0^{+}}\frac{\ln(5x)}{\frac{1}{3x}}$, which is in the $\frac{\infty}{\infty}$ form.

Step2: Apply L'Hopital's rule

Differentiate the numerator and denominator. The derivative of $\ln(5x)$ is $\frac{1}{x}$, and the derivative of $\frac{1}{3x}=-\frac{1}{3x^{2}}$. So, $\lim_{x
ightarrow0^{+}}\frac{\ln(5x)}{\frac{1}{3x}}=\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{-\frac{1}{3x^{2}}}=\lim_{x
ightarrow0^{+}}(- 3x)=0$.

Step3: Find the original limit

Since $\lim_{x
ightarrow0^{+}}\ln y = 0$, and $y = e^{\ln y}$, then $\lim_{x
ightarrow0^{+}}y=\lim_{x
ightarrow0^{+}}(5x)^{3x}=e^{0}=1$.

Answer:

$1$