Evaluating a $1^{\infty}$ limit using logarithms and L'Hopital's Rule

Let $y = \left(1+\frac{3}{x}\right)^x$. Taking the natural log gives $\ln y = x\ln\left(1+\frac{3}{x}\right)$, which can be rewritten as $\ln y = \dfrac{\ln(1+3/x)}{1/x}$. As $x\to\infty$, the numerator $\ln(1+3/x)\to \ln(1)=0$ and the denominator $1/x \to 0$, so this is the indeterminate form $0/0$, allowing L'Hopital's Rule to be applied.

Differentiating numerator and denominator separately with respect to $x$: the derivative of $\ln(1+3/x)$ is $\dfrac{-3/x^2}{1+3/x}$, and the derivative of $1/x$ is $-1/x^2$. Dividing gives $\dfrac{-3/x^2}{1+3/x} \cdot \dfrac{x^2}{-1} = \dfrac{3}{1+3/x}$, and taking the limit as $x\to\infty$ gives $\lim_{x\to\infty} \ln y = 3$.

Since $\ln y \to 3$, exponentiating both sides gives $y \to e^3$. Therefore $\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^x = e^3.$